A 10-kg block with an initial velocity of 10 m/s slides 10 m across a horizontal surface and comes to rest. It takes the block 2 s to stop. The stopping force acting on the block is about
A)5 N.
B)10 N.
C)25 N.
D)50 N.
E)none of the above

Question

Quizplus · Accepted Answer

The stopping force can be calculated using the formula $F = \frac{\Delta p}{\Delta t}$, where $\Delta p$ is the change in momentum and $\Delta t$ is the change in time. The initial momentum is $p_i = mv = 10 \times 10 = 100 \, \text{kg m/s}$, and the final momentum is $0 \, \text{kg m/s}$ since the block comes to rest. Thus, $\Delta p = 0 - 100 = -100 \, \text{kg m/s}$, and $\Delta t = 2 \, \text{s}$. Substituting these values gives $F = \frac{-100}{2} = -50 \, \text{N}$. The negative sign indicates the direction of the force opposite to the motion, but the magnitude of the force is $50 \, \text{N}$.

A 10-Kg Block with an Initial Velocity of 10 M/s