You apply a 50 n rightward force to a 4 kg cart to accelerate it across a horizontal surface at a rate of 3.28m/s2.

determine the force of friction acting upon the cart.
determine the coefficient of friction acting on the cart.

The force of friction is 36.88 N

The coefficient of friction is 0.94

Explanation:

Lets explain how to solve the problem

The force applied is 50 N rightward

The mass of the cart is 4 kg

The applied force makes the cart accelerate across a horizontal surface

The acceleration is 3.28 m/s²

We need to find the force of friction acting upon the cart

According to Newton's law

→ ∑ F in direction of motion = mass × acceleration

We have two horizontal forces acting on the cart

→ The force applied = 50 N

→ The force of friction in opposite direction of motion

→ The mass = 4 kg

→ Acceleration = 3.28 m/s²

Substitute these values in the rule

→ 50 - = 4 × 3.28

→ 50 -   = 13.12

Add   for both sides

→ 50 =   + 13.12

Subtract 13.12 from both sides

→ 36.88 =  

The force of friction is 36.88 N

→ The force of friction = μ R

where R is the normal reaction and μ is the coefficient of friction

→ R = mg

where m is the mass of the cart and g is the acceleration of gravity

→ = μ mg

→ g = 9.8 m/s² , m = 4 kg , = 36.88 N

Substitute these values in the rule

→ 36.88 = μ (4)(9.8)

→ 36.88 = 39.2 μ

Divide both sides by 39.2

→ μ = 0.94

The coefficient of friction is 0.94

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

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