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A 3.0-kg block moves up a 40° incline with constant speed under the action of a 26-N force acting up and parallel to the incline. What magnitude force must act up and parallel to the incline for the block to move down the incline at constant velocity?

Respuesta :

brightediomo1

Explanation:

Sooooo, first,

Constant Speed means that the body isn't accelerating.

a= 0

Let the force that causes it to move up the incline be P and Q be the force that causes the body to move down the incline. Let Fr be the frictional force.

The force that pulls the body downwards is given by,

F = mgsin40°

F= 18.898N

P -(F+Fr) = ma, but, a= 0

P = F + Fr

Fr = 26-18.898

= 7.102N

If the block must move down the incline, Q and Fr will act in the same direction.

Q + Fr = F

Q = 11.796N

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