Question

A 80 kg block starts from rest at the top of a 15m frictionless ramp. The block slides down the frictionless ramp onto a rough horizontal surface, where it eventually comes to rest 4m later. How much total work is done by the force of friction on the block as it comes to rest?

Answer 1

Answer:

W(fric) = -12000 J

Explanation:

Given that

mass ,m= 80 kg

Initial speed ,u= 0 m/s

Height ,h= 15 m

Final speed ,v= 0 m/s

We know that

Work done by all the forces = Change in the kinetic energy

$W_{gravity}+W_{fric}=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

$W_{gravity}+W_{fric}=\dfrac{1}{2}m\times 0^2-\dfrac{1}{2}m\times 0^2$

$W_{gravity}+W_{fric}=0$

W(fric)= - m g h

W(fric) = - 80 x 10 x 15     ( g=10 m/s²)

W(fric) = -12000 J

negative sign indicates that force and displacement is in opposite direction.