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3 years ago

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A 30.0 kg packing case is initially at rest on the floor of a 1500 kg pickup truck. The coefficient of static friction between t

he case and the truck floor is 0.30, and the coefficient of kinetic friction is 0.20. Before each acceleration given below, the truck is traveling due north at constant speed. Find the magnitude and direction of the friction force acting on the case (a) when the truck accelerates at 2.20 m/s2 northward and (b) when it accelerates at 3.40 m/s2 southward.

1 answer:

alexgriva [62]3 years ago

7 0

Answer:

(a). The magnitude of the frictional force acting on the case is 66 N northward.

(b). The magnitude of the friction force acting on the case is 58.8 N southward.

Explanation:

Given that,

Mass of packing case = 30.0 kg

Mass of truck = 1500 kg

Coefficient of static friction = 0.30

Coefficient of kinetic friction = 0.20

Acceleration = 2.20 m/s² northward

(A). We need to calculate the force acting on the case

Using formula of force

$F=ma$

Put the value into the formula

$F=30.0\times2.20$

$F=66\ N$

We need to calculate the maximum static frictional force

Using formula of frictional force

$f_{\mu}=\mu mg$

Put the value into the formula

$F_{\mu}=0.30\times30.0\times9.8$

$F_{\mu}=88.2$

The frictional force is greater then the force acting on the block

Then, The frictional force is 66 N northward.

(B). We need to calculate the force acting on the case

Using formula of force

$F = ma$

Put the value into the formula

$F=30.0\times3.40$

$F=102\ N$

The static friction force is less then the force action on the case

We need to calculate the maximum static frictional force

Using formula of frictional force

$f_{\mu}=\mu mg$

$f_{\mu}=0.20\times30.0\times9.8$

$f_{\mu}=58.8\ N$

So, The magnitude of the friction force acting on the case is 58.8 N southward.

Hence, This is the required solution.

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