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A force of 46N is applied horizontally to a box with a mass of 0.5kg. The block sits on a horizontal table where the coefficient of friction between the box and the table is .5.What is the acceleration of the box? (In m/s^2)

Question

A force of 46N is applied horizontally to a box with a mass of 0.5kg. The block sits on a horizontal table where the coefficient of friction between the box and the table is .5.

What is the acceleration of the box? (In m/s^2)

check_circleAnswer
Step 1

Given:

Mass of the object = 0.5 kg

Coefficient of friction = 0.5

Value of applied force = 46 N

Step 2

Calculating the accelerat...

Net force on the object, F= Applied force - Friction force
net
Friction force, Friction = fiN = mg
Subtuting the values, iion (0.5)0.5kg)(9.8m/s2)
Friction
F,
Friction = 2.45N
Hence, F 46N-2.45N
net
F43.55N
net
43.55N 7.1m/s
Net aceleration
ner
0.5kg
т
a 87.1m/
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Net force on the object, F= Applied force - Friction force net Friction force, Friction = fiN = mg Subtuting the values, iion (0.5)0.5kg)(9.8m/s2) Friction F, Friction = 2.45N Hence, F 46N-2.45N net F43.55N net 43.55N 7.1m/s Net aceleration ner 0.5kg т a 87.1m/

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Physics

Newtons Laws of Motion

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