An 103 Newtons box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at an acceleration of -0.9m/s2. The push force has a horizontal component of 20N and a vertical component of 25N downward. Calculate the coefficient of the kinetic friction and the box. Hint. Draw all the forces. Apply Newtons 2nd law : ƩFy to solve for the normal force and ƩFx to solve for the frictional force. Then apply the formula : frictional force = µkN.

asked by job

8 years ago

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M*g = 103 N. = Wt. of box.
M = 103/g = 103/9.8 = 10.5 kg.

Fn = 103*Cos 0 + 25 = 128 N. = normal force.

Fk = u*Fn = 128u.

Fx-Fk = M*a.
20-128u = 10.5*(-0.9), u = ?.

answered by Henry

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Question

An 103 Newtons box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at an acceleration of -0.9m/s2. The push force has a horizontal component of 20N and a vertical component of 25N downward. Calculate the coefficient of the kinetic friction and the box. Hint. Draw all the forces. Apply Newtons 2nd law : ƩFy to solve for the normal force and ƩFx to solve for the frictional force. Then apply the formula : frictional force = µkN.

1 answer

M*g = 103 N. = Wt. of box.
M = 103/g = 103/9.8 = 10.5 kg.

Fn = 103*Cos 0 + 25 = 128 N. = normal force.

Fk = u*Fn = 128u.

Fx-Fk = M*a.
20-128u = 10.5*(-0.9), u = ?.

Ask a New Question or answer this question.

Henry · Answer

M*g = 103 N. = Wt. of box. M = 103/g = 103/9.8 = 10.5 kg. Fn = 103*Cos 0 + 25 = 128 N. = normal force. Fk = u*Fn = 128u. Fx-Fk = M*a. 20-128u = 10.5*(-0.9), u = ?.