Suppose a box is being dragged by a force of 100 N over a rough surface. If the

Question

Suppose a box is being dragged by a force of 100 N over a rough surface. If the coefficient of kinetic friction at the interface of the box and surface is 0.2, and the box is moving with a constant speed of 2 m/s, calculate the net work done on the box if it is dragged for 1.0 m.

A. 0 J

B. 100 J

C. 200 J

D. 50 J

E. 20 J

A cab driver suddenly slams on the brakes, and skids a certain distance on a straight level road. If you had been traveling twice as fast, what distance would the car have skidded, under the same conditions?
A. It would have skidded 4 times farther.
B. It would have skidded twice as far.
C. It would have skidded 1.4 times farther.
D. It would have skidded one half as far.
E. It is impossible to tell from the information given.

The momentum of an isolated system is conserved
A. only in inelastic collisions.
B. only in elastic collisions.
C. in both elastic and inelastic collisions.

Explanation / Answer

part 1:

as the box is moving with constant speed, net force on the block is zero

hence net work done=force*distance=0 J

hence option A is correct.

part 2:

as we know that, final speed^2-initial speed^2=2*acceleration*distance

here initial speed=u (let , for the cab driver)

final speed=0

let acceleration be -a .(where a>0, acceleration is negative because it opposes the motion)

hence 0^2-u^2=-2*a*distance

==>distance=u^2/(2*a)

with acceleration remaining constant at -a,

initial speed becomes 2*u.

then applying the same formula:

0^2-(2*u)^2=-2*a*distance

==>distance=4*u^2/(2*a)

which is 4 times the distance skidded earlier.

hence option A is correct.

part 3:

as isolated system means there is no external force acting on it,

total momentum is conserved in case of both elastic and inelastic collision.

where as for elastic collision, total energy is conserved but for inealstic collison total energy is not conserved.

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