A 12.5kg crate slides along a horizontal frictionless surface at a constant spee

Question

A 12.5kg crate slides along a horizontal frictionless surface at a constant speed of 4.0m/s. The crate then slides down a frictionless incline and across a second horizontal surface. 1: while the crate slides along the upper surface, how much gravitational potential energy does it have compared to what it would have on the lower surface?A: 80J B: 100J C:240J D: 370J 2: what is the speed of the crate when it arrives at the lower surface? A: 7.7m/s B:8.6m/s C:59m/s D: 75m/s 3: what minimum coefficient of kinetic friction is required to bring the crate to a stop over a distance of 5.0m along the lower surface? A 0.30 B: 0.60 C: 0.76 D: 0.82

Explanation / Answer

You have missed out part of the question - what is the height reduction (h) when the crate slides down the incline? The formula fr kinetic energy is KE = mv^2/2 Initial KE = 12.5 x (4.0)^2/2 = 100J This will increase by the potential energy lost, mgh. Just work out mgh and add to 100J. mgh should work out to 370J. To stop the mass, the work done by the force must equal the kinetic energy: WD =F x d 470 = F x 5 F = 470/5 = 94m

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