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A force of 400 Newton acts on a body of mass 100 kg is resting on a surface whose coefficient of friction is 0.3 the acceleration is?
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Given:
Force (F) = 400 N
Mass (m) = 100 kg
Coefficient of friction (μ) = 0.3

Calculating the force of friction:
The force of friction can be calculated using the equation:
Frictional force (Ff) = μ * Normal force (N)

The normal force is the force exerted by the surface on the body and is equal to the weight of the body. The weight can be calculated using the equation:
Weight (W) = mass (m) * acceleration due to gravity (g)

Assuming the acceleration due to gravity is 9.8 m/s², we can calculate the weight:
Weight (W) = 100 kg * 9.8 m/s² = 980 N

Therefore, the normal force (N) = 980 N

Now, we can calculate the force of friction:
Ff = 0.3 * 980 N = 294 N

Calculating the net force:
The net force acting on the body is the difference between the applied force and the force of friction:
Net force (Fnet) = Force (F) - Frictional force (Ff)

Fnet = 400 N - 294 N = 106 N

Calculating the acceleration:
Using Newton's second law of motion, we know that the net force is equal to the mass multiplied by the acceleration:
Fnet = m * a

106 N = 100 kg * a

Solving for acceleration (a):
a = 106 N / 100 kg = 1.06 m/s²

Conclusion:
The acceleration of the body is 1.06 m/s².
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A force of 400 Newton acts on a body of mass 100 kg is resting on a surface whose coefficient of friction is 0.3 the acceleration is?
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