Question

600 kg elephant runs at 5 m/s and jumpts onto a 100kg cart. If the coefficient of friction is .04 how far will the cart travel on a horizontal surface before coming to rest?

Answer 1

Answer:

Eleven seconds.

Explanation:

Two keys are needed to solve this problem. First, the conservation of momentum: allowing you to calculate the cart's speed after the elephant jumped onto it. It holds that:

$m_ev_e+m_c\cdot0=(m_e+m_c)v_0\implies \\v_0=\frac{m_ev_e}{m_e+m_c}=\frac{600kg\cdot 5\frac{m}{s}}{700kg}=4.29\frac{m}{s}$

So, once loaded with an elephant, the cart was moving with a speed of 4.29m/s.

The second key is the kinematic equation for accelerated motion. There is one force acting on the cart, namely friction. The friction acts in the opposite direction to the horizontal direction of the velocity v0, its magnitude and the corresponding deceleration are:

$F_r = 0.04\cdot (m_e+m_c)g\implies a_r = 0.04\cdot g= 0.04 \cdot 9.8 \frac{m}{s^2}=0.39\frac{m}{s^2}$

The kinematic equation describing the decelerated motion is:

$v = -a_r t+v_0\\0 = -a_rt+v_0\implies \\t = \frac{v_0}{a_r}=\frac{4.29\frac{m}{s}}{0.39\frac{m}{s^2}}=11s$

It takes 11 seconds for the comical elephant-cart system to come to a halt.