CH 6-9

25 July 2022
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The SI unit for work
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Joule (J) 1J=1Nm=1(kgm2/s2)
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The net work done on an object is
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the sum of the work done by each individual force acting on that object. In other words, Wnet=W1+W2+W3+?=?iWi .
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The net work can also be expressed as the work done by the net force acting on an object, which can be represented by the following equation:
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Wnet=Fnet d cos?
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Calculating the kinetic energy
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1/2 mv^2
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Calculating the potential energy
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PE = m*g*h (mgh)
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Calculating Work (W)
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W = F? ?x W = F * ?r (work done by constant force moving an object through a straight-line displacement)
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Calculating fricitonal force
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fk = µ?n
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Calculating avg Power
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P= Fs/t = Fvavg.
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Finding the amount of work done by a constant force
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W = Fd cos theta
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Work and the scalar product
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W = (F cos ?) (?r) = = F ?r cos ? A . B = AB cos ?
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Dot product
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A.B = B.A A.(B+C) = A.B+A.C A . B = AxBx + AyBy + AzBz
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Work done by a varying voce in one dimension
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W = ??² ?? F(x) dx
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??² ?? x? dx =
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(x??¹)/(n+1) |?² ?? = = (x???¹)/(n+1) - (x???¹)/(n+1)
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W = ??? F(x) dx =
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??? kx dx = 1/2kx²|?? = 1/2k(x)² - 1/2k(0)² = = 1/2kx²
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Force and Work in two and three dimensions
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??²?? F *dr
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Wnet =
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? Fnet dx = ?m(dv/dt)dx = ?m dv (dx/dt) = ????² mv dv = = 1/2mv²|???² = = 1/2mv?² - 1/2mv?²
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Wspring=
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?(1/2)kx^2
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Work-energy theorem
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?K = Wnet
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average power
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P = (?W/?t)
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instantaneous power
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P = lim(?t?0) (?W/?t) = = dW/dt
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Amount of work done in time
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W = P?t
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Amount of work done in time when power is not constant
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W = lim(?t?0) ? P ?t = = ??² P dt
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Power and Velocity
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dW = F * dr P = (dW/dt) = F*(dr/dt) P = F*v
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Two blocks of ice, one four times as heavy as the other, are at rest on a frozen lake. A person pushes each block the same distance d. Ignore friction and assume that an equal force F? is exerted on each block. What does this tell you?
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It is equal to the kinetic energy of the lighter block.
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Compared to the speed of the heavier block, what is the speed of the light block after both blocks move the same distance d?
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twice as fast
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Now assume that both blocks have the same speed after being pushed with the same force F? . What can be said about the distances the two blocks are pushed?
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The heavy block must be pushed 4 times farther than the light block.
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Law of the conservation of energy
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Ki + Ui = Kf + Uf
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Let vectors A? =(2,1,?4), B? =(?3,0,1), and C? =(?1,?1,2). Calculate A? ?B
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-10
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Let vectors A? =(2,1,?4), B? =(?3,0,1), and C? =(?1,?1,2). What is the angle ?AB between A? and B? ?
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2 radians
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Let vectors A? =(2,1,?4), B? =(?3,0,1), and C? =(?1,?1,2). Calculate 2B? ?3C
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30
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Let vectors A? =(2,1,?4), B? =(?3,0,1), and C? =(?1,?1,2). Calculate 2(B? ?3C? )
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30
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V? 1 and V? 2 are different vectors with lengths V1 and V2 respectively. V? 1?V? 1 =
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V12
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If V? 1 and V? 2 are perpendicular,
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V? 1?V? 2 = 0
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If V? 1 and V? 2 are parallel
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V? 1?V? 2 = V1V2
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consider a sled of mass m being pulled by a constant, horizontal force of magnitude F along a rough, horizontal surface. The sled is speeding up. How many forces are acting on the sled?
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4
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The work done on the sled by the normal force is
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also 0
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The work done on the sled by the pulling force is (positive or negative?)
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positive
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The work done on the sled by the force of friction is (positive or negative?)
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negative
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The net work done on the sled is pos or neg?
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pos
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Does the kinetic energy of the sled increase, decrease or remains constant?
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increases
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The work-energy theorem states that
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the change in kinetic energy of an object equals the net work done on that object.
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the initial kinetic energy of each block is zero, both blocks have
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the same final kinetic energy.
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consider a sled of mass m being pulled by a constant, horizontal force of magnitude F along a rough, horizontal surface. The sled is speeding up. How would we find the net work done on the sled?
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Determine which force is bigger; in this case, The magnitude of the pulling force is greater than that of the force of friction. Therefore, the net work is positive
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In general, the work done by a constant force F? can be found as
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W=Fscos(?)
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? is the angle between vectors F? and s? . However, when the net force and displacement have the same direction (as is the case here), cos(?)=
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1
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You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,500 N across the carpet to a spot 5 m away on the opposite wall. Hoping to just slide your dresser easily across the floor, you do not empty your clothes out of the drawers before trying to move it. You push with all your might but cannot move the dresser before becoming completely exhausted. How much work do you do on the dresser?
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W = zero Joules
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The magnitude of the force of attraction between two objects of masses M1 and M2 that are separated by a distance r is given by:
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F=G (M1M2/r2).
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The circumference of a circle with radius r is
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C=2?r
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If an object undergoes displacement while being acted upon by a force (or several forces), it is said that work is being done on the object. If the object is moving in a straight line and the displacement and the force are known, the work done by the force can be calculated as W=
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W=F? ?s? =??F? ????s? ??cos?
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If the speed is constant, the force is equal to
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the frictional force Ffriction = uk*mg
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How do yoyu find the amount of energy E dissipated by friction by the time an object stops?
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Kinetic energy + potential energy K + U
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The work done by frictional force is always positive or negative?
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negative
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What does the conservation of energy tell us when the potential energy is lost?
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It must be converted to kinetic energy
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F|| is the component of F? that is either parallel or antiparallel to the displacement. If F|| is parallel to d? , as in the case of F? p, then the work done is (pos/neg?). If F|| is antiparallel to d? , as in the case of f? k, then the work done is (pos/neg?).
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positive; negative
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the x coordinate xcm of the center of mass of the system.
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xcm = m1x1+m2x2/m1+m2
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If m2?m1, then the center of mass is located
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o the left of m2 at a distance much less than x2?x1
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If m2=m1, then the center of mass is located:
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half-way between m1 and n2
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Recall that the blocks can only move along the x axis. The x components of their velocities at a certain moment are v1x and v2x. Find the x component of the velocity of the center of mass (vcm)x at that moment. Keep in mind that, in general: vx=dx/dt.
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(vcm)x = v1xm1+v2xm2/m1+m2
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Suppose that v1x and v2x have equal magnitudes. Also, v1? is directed to the right and v2? is directed to the left. The velocity of the center of mass is then:
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he answer depends on the ratio m1/m2
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Assume that the x components of the blocks' momenta at a certain moment are p1x and p2x. Find the x component of the velocity of the center of mass (vcm)x at that moment.
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|p1x|=|p2x|
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Assume that the blocks are accelerating, and the x components of their accelerations at a certain moment are a1x and a2x. Find the x component of the acceleration of the center of mass (acm)x at that moment. Keep in mind that, in general, ax=dvx/dt.
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(acm)x = a1xm1+a2xm2/m1+m2
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Consider the same system of two blocks. An external force F? is now acting on block m1. No forces are applied to block m2 as shown (Figure 2) . Find the x component of the acceleration of the center of mass (acm)x of the system.
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(acm)x = Fx/m1+m2
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Consider the same system of two blocks. Now, there are two forces involved. An external force F1? is acting on block m1 and another external force F2? is acting on block m2. (Figure 3) Find the x component of the acceleration of the center of mass (acm)x of the system.
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(acm)x = F1x+F2x / m1+m2
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Consider the previous situation. Under what condition would the acceleration of the center of mass be zero? Keep in mind that F1x and F2x represent the components, of the corresponding forces.
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F1x=?F2x
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Consider the same system of two blocks. Now, there are two internal forces involved. An internal force F? 12 is applied to block m1 by block m2 and another internal force F? 21 is applied to block m2 by block m1 (Figure 4) . Find the x component of the acceleration of the center of mass (acm)x of the system.
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(acm)x = F12x+F21x / m1+m2