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  1. A 2 kg block slides down a rough incline with a constant velocity. The angle

of incline is θ=30⁰. Coefficient of friction of the incline is, μk=0.5. A

horizontal force, F, is acting on the block as shown in the figure. (Assume

g=10 m/s and do not forget that the direction is downward!!) (Note: You

don’t have to use unit vectors. Mentioning the sign convention correctly for

all forces is enough for the answer.)

a) Determine the magnitude of the force F.

b) Determine the normal force exerted by the incline surface to the block

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kevin1723

Lvl 2
2y ago
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Kathlyn Feil

Lvl 1
2y ago
great answer, thanks!
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Wiki User

11y ago

s=s^i+v^i*t+ 1/2 at^2

As block is starting from rest so its s^i and v^i will be zero,,

now eqn comes to s=1/2 at^2

..

by putting the values u can easiily find the answer.!

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Wiki User

12y ago

displacement = (avg velocity /2) * t

2m = (avg velocity /2) * 1.5 sec

avg velocity = 2.67m/s

v/t= a

2.67/1.5s = 1.78m/s2

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Q: A 3.70 kg block starts from rest at the top of a 30.0 degree incline and slides 2 m down the incline in 1.50 s Find the acceleration of the block?
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