Question

29. A 5.0-kg cart is moving horizontally at 5m/s. In order to change its speed to 10m/s, the
net work done on the cart must be:
A. 40 J B. 190j C. 160 J D. 100)​

Answer 1

Answer:

Explanation:

INITIAL KE = $\frac{1}{2}mv^{2}$

                  = $\frac{1}{2}5^{3}$

                 = $\frac{125}{2}$

FINAL KE = $\frac{1}{2}mv^{2}$

               = $\frac{1}{2}X5 X10^{2}$

             = 50x5=250

CHANGE IN KE = 250-125/2

                          = (500-125)/2

                         = 375/2J