Question

A 1500 kg car begins sliding down a 5.0o inclined road with a speed of 30 km/h. The engine is turned off, and the only forces acting on the car are a net frictional force from the road and the gravitational force. After the car has traveled 50 m along the road, its speed is 40 km/h.
(a) How much is the mechanical energy of the car reduced because of the net frictional force?
(b) What is the magnitude of that net frictional force?

Answer 1

Answer:

a) -23583J

b) 471.66N

Explanation:

Using the conservation energy theorem:

$K_o+U_o+W_f=K_f+U_f\\W_f=K_f+U_f-(K_o+U_o)$

The height is given by:

$sin(\theta)=\frac{h}{d}\\\\h=50*sin(5^o)\\h=4.36m$

substituting the values we have:

$W_f=\frac{1}{2}*1500kg*((40-30)\frac{km}{h}*\frac{1000h.m}{3600km.s})^2+0-1500kg*9.8m/s^2*4.36m\\W_f=-23583J$*

the work is defined by:

$W_f=F_f*d*cos(\theta)$

the force of the friction force is 180 degrees because it's opposite to the movement, so:

$F_f=\frac{-23583J}{50m*cos(180)}\\F_f=471.66N$