N=mgcos@=281cos45
Ffr = u * N = .16 * 281cos45
Wfr = Ffr*d = Ffr * 6.2
Work done by friction is the energy transfered to thermal, and is your change in total energy.
Thus for part b, E0 = mgh + mv^2/2
at the botto she has E0 - Wfr total energy and it is all kinetic.
From there do the algebra and sub in numbers.
A child whose weight is 281 N slides down a 6.20 m playground slide that makes an angle of 45.0¡ã 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 0.516 m/s, what is her speed at the bottom?
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