The kinetic friction force that a horizontal surface exerts on a 60.0-kg object is 50.0 N.If the initial speed of the object is 25.0 m/s,what distance will it slide before coming to a stop?
A)15.0 m
B)30.0 m
C)375 m
D)750 m

Question

Quizplus · Accepted Answer

We can use the kinetic energy equation to solve for the distance:
KE = 1/2 mv^2
When the object comes to a stop, all the kinetic energy is converted to work done by friction:
KE = work = force x distance
Therefore, we can set these two equations equal to each other:
1/2 mv^2 = Fd
Solving for d, we get:
d = (1/2 mv^2)/F
Plugging in the given values, we get:
d = (1/2 x 60.0 kg x (25.0 m/s)^2)/50.0 N
d = 375 m
Therefore, the answer is C) 375 m.

The Kinetic Friction Force That a Horizontal Surface Exerts on a 60.0-Kg