3G .dl 16%7:04 PM A 4.60 kg block located on a horizontal frictionless floor is

Question

3G .dl 16%7:04 PM A 4.60 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F-12.5N at an angle theta-25.5degrees above the horizontal, as shown. What is the magnitude of the acceleration of the block when the force is applied? Enviar Respuesta Tries 0/10 What is the horizontal speed of the block 4.90 seconds after it starts moving? Enviar Respuesta Tries 0/10 block when the force F is acting on it? Enviar Respuesta Tries 0/10 If, instead, the floor has a coefficient of kinetic friction 0.04, what is the magnitude of the frictional force on the block when the block is moving? Enviar Respuesta Tries 0/10 What is the magnitude of the acceleration of the block when friction is being considered?

Explanation / Answer

Reolve the applied force into two component, as block is moving on a horizontal floor, so we will consider only horizontal component

F = 12.5*cos 25.5

F = 11.2823 N

Now, we know that

F = ma

Therefore,

a = F/m

a = 11.2823 / 4.60

a = 2.4526 m/s2

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Now, horizontal speed after 4.9 sec

v = u +at

v = 0 + 2.4526*4.90

v = 12.018 m/s

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Normal force is the force acting perpendicula to the surface

Fnormal = mg - F*sin25.5

Fnormal = 39.69 N

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Frictional force

It acts in the direction opposite of the direction of motion

Therefore,

Fnet = ma

12.5*cos 25.5 - Ffriction = 4.60*2.4526

solve for Ffriction

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