Question

A child whose weight is 281 N slides down a 6.10 m playground slide that makes an angle of 47.0° with the horizontal. The coefficient of kinetic friction between slide and child is 0.160. (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of 0.311 m/s, what is her speed at the bottom?

Answer 1

the friction force is

The normal force on an incline is given by

weight W = mg = 281 N

Work done by the friction force is negative as it acts opposite to the displacement

W_f = fd = -30.66 (6.1) = -187 J

the energy is transferred to thermal energy is the work done by friction i.e 187 J

b) initial speed at top v₁ = 0.311 m/s

W = mg = 281 N

m = 281/9.81

m = 28.64 Kg

initial kinetic energy

Work done by gravity force is, mg sin theta is parallel to the displacement so work is positive, mg cos theta is perpendicular to displacement so work is zero due to mg cos theta

using work-energy theorem, change in kinetic energy is equal to total work done

her speed at the bottom is 8.64 m/s