Velocity-time graph for a body of mass 10 kg is shown in figure. Work-done on the body in first two seconds of the motion is :

A car of weight W is on an inclined road that rises by 100 m over a distance of 1 km and applies a constant frictional force $${W \over 20}$$ on the car. While moving uphill on the road at a speed of 10 ms⁻¹, the car needs power P. If it needs power $${p \over 2}$$ while moving downhill at speed v then value of $$\upsilon $$ is :

A person trying to lose weight by burning fat lifts a mass of $$10$$ $$kg$$ upto a height of $$1$$ $$m$$ $$1000$$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $$3.8 \times {10^7}J$$ of energy per $$kg$$ which is converted to mechanical energy with a $$20\% $$ efficiency rate. Take $$g = 9.8\,m{s^{ - 2}}$$ :

A point particle of mass $$m,$$ moves long the uniformly rough track $$PQR$$ as shown in the figure. The coefficient of friction, between the particle and the rough track equals $$\mu .$$ The particle is released, from rest from the point $$P$$ and it comes to rest at point $$R.$$ The energies, lost by the ball, over the parts, $$PQ$$ and $$QR$$, of the track, are equal to each other , and no energy is lost when particle changes direction from $$PQ$$ to $$QR$$.

The value of the coefficient of friction $$\mu $$ and the distance $$x$$ $$(=QR),$$ are, respectively close to: