Question

A
10 kg block is launched up a plane inclined at a 15° angle. The
initial speed of the block is 5 m/s.

a) Using Newton's laws of motion and the equations of
kinematics, calculate how far up the inclined plane does the block
slide in the absence of friction?

b) Using work and energy, answer the question in part (a) in
the presence of friction, taking the coefficient of kinetic
friction between the block and the surface to be µk=.2

c) [2 points] Let the coefficient of static friction between
the block and the surface be µs=0.35. In the scenario in part (b),
once the block stops, does it start to slide back down or does it
remain at rest?EXPLAIN

Answer #1

a] Initial speed = 5 m/s

acceleration a = - gsin15 = - 2.536 m/s^{2}

use v^{2} = u^{2} + 2aL

at the final position v = 0 m/s

0 = 5^{2} + 2( - 2.536)L

=> L = 4.928 m

b]

initial kinetic energy = (1/2)mu^{2}

at the highest point, the kinetic energy is zero.

so, using conservation of energy,

(1/2)mu^{2} = mgLsin15 + u_{k}(mgcos15)L

=> (1/2)u^{2} = gLsin15 + u_{k}(gcos15)L

=> L = 2.82 m.

c] u_{s} = 0.35

the net force on the block is:

F = mgsin15 - u_{s}mgcos15 = - 7.76 N

which is negative. Therefore, the block remains at rest.

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