Posted by Sandhya on Sunday, June 13, 2010 at 10:59pm.
(a)
Distance, S = 9m
initial velocity, u = 0 m s-2
final velocity, v = 2.4 m s-2
acceleration, a (assumed uniform)
= (v²-u²)/2S
= (2.4²-0²)/(2*9)
= 0.32 ms-2
Acceleration due to gravity, g
= 9.8 m s-2
component of g along incline
= g sin(θ)
= 9.8 sin(15°)
= 2.536 m s-2
Therefore reduced acceleration due to friction, af
= 2.536-0.32
= 2.216 ms-2
Frictional force, F
= m(af)
= 25 kg * 2.216 m s-2
= 55.411 N
Normal force acting on inclined plane, N
= m * g cos(θ)
= 25 * 9.8 * cos(15°)
= 236.6 N
Coefficient of kinetic friction, μ
= F/N = 0.234
(b) work done = force * distance
(c) Change in potential energy = mgh
where h is the difference in elevation (final - initial).
if the mass were suspended by water and it's specific gravity were less than one would it knot float? slip up!
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