17. In the figure, a force F is being applied to the block with a magnitude such that it moves up along the frictionless inclined plane with a constant velocity of 5.0 m/s. The angle in the figure is 30Ā°, and the mass of the block is 50 kg. How much work (in Joules) is done by the force F in moving the block 10 meters along the surface of the inclined plane?
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The angle of the incline is 30 degrees, and the force F is applied parallel to the incline. You need to find the work done by the force F as the block moves 10 meters along the incline. Show moreā¦
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Nishant K.
In Fig. $8-51$ , a block slides down an incline. As it moves from point $A$ to point $B,$ which are 5.0 $\mathrm{m}$ apart, force $\vec{F}$ acts on the block, with magnitude 2.0 $\mathrm{N}$ and directed down the incline. The magnitude of the frictional force acting on the block is 10 $\mathrm{N} .$ If the kinetic energy of the block increases by 35 $\mathrm{J}$ between $A$ and $B$ , how much work is done on the block by the gravitational force as the block moves from $A$ to $B ?$
A 10.0 kg block is set into motion up an inclined plane with an initial speed of 8.5 m/s, as shown in the Figure. The block comes to rest after traveling 3.0 m along the plane, which is inclined at an angle of 30Ā° to the horizontal. The kinetic coefficient of friction Ī¼k between the block and the inclined plane is a) 0.37 b) 0.12 c) 0.51 d) 0.24 e) 0.81 e) 0.65. vi = 8.5 m/s Hint: Consider g = 10.0 m s2.
Ankur S.
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