Answer to Question #169759 in Mechanics | Relativity for Mae

Question

A crate weighing 80N is pulled along a horizontal surface by a force of 15 N which is applied 20° above the horizontal. If the crate from rest and the coefficient of friction is 0.15, determine the kinetic energy and the velocity of the crate after it has moved a distance of 10 m.

Answer 1

• Refer to the sketch attached

• Apply F=ma vertically,

$\qquad\qquad \begin{aligned} \small F&= \small ma\\ \small R+15\sin20-80&= \small0\\ &= \small 80-5.13\\ \small R&= \small 74.87N \end{aligned}$

• As it moves under friction, it means it is experiencing the critical static friction, then,

$\qquad\qquad \begin{aligned} \small f&= \small \mu R\\&= \small 0.15\times74.87\\ &= \small11.23N \end{aligned}$

• Mass of the block

$\qquad\qquad \begin{aligned} \small m&= \small \frac{80N}{9.8ms^{-2}}\\ &= \small 8.16kg \end{aligned}$

• Apply F=ma to the moving direction to calculate the acceleration of the block,

$\qquad\qquad \begin{aligned} \small \to F&= \small ma\\ \small 15\cos20-11.23&= \small 8.16kg\times a\\ \small a&= \small \frac{2.865}{8.16}\\ &= \small 0.35ms^{-2} \end{aligned}$

• Since it starts from rest, apply an appropriate equation of the 4 equations of motion.

$\qquad\qquad \begin{aligned} \small \to V^2&= \small U^2+2as \\ \small V^2&= \small 0+2\times 0.35ms^{-2}\times10m\\ \small V^2&= \small 7m^2s^{-2}\cdots(1)\\ \small V&= \small \bold{2.65ms^{-1}} \end{aligned}$

• Kinetic energy it has at 10m is

$\qquad\qquad \begin{aligned} \small E_k&= \small \frac{1}{2}mV^2\\ &= \small \frac{1}{2}\times 8.16kg\times 7m^2s^{-2}\\ &= \small \bold{28.56\,J} \end{aligned}$