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Rasek [7]
3 years ago
8

A 30.0-kg packing case is initially at rest on the floor of a1500-kg pickup truck. The coefficient of static friction betweenthe

case and the truck floor is 0.30 and the coefficient of kineticfriction is 0.20. Before each acceleration given below, the truckis traveling due north at constant speed.(a) Find the magnitude and direction of thefriction force acting on the case when the truck accelerates at2.51 m/s2 northward.magnitude Ndirection (b) Find the magnitude and direction of the friction force actingon the case when the truck accelerates at 3.63 m/s2 southward.magnitude Ndirection
Physics
1 answer:
777dan777 [17]3 years ago
5 0

Answer:

Explanation:

a ) maximum friction possible

= .3 x 30 x 9.8

= 88.2 N

It is friction force which creates acceleration in 30 kg packing case.

Friction force F

F = ma

= 30 x 2.51

= 75.3 N

It will be in north direction , the direction of acceleration.

b )  F = ma

= 30 x 3.63

= 108.9 N

But maximum friction force is 75.3 N , so load will start slipping northward. so friction force will be acting southward.

Friction force = .2 x 30 x 9.8 ( coefficient of kinetic friction applies )

= 58.8 N

towards south .

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Answer:

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********************

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Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\

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