Question

As shown in the figure above, a 50 kg box is dragged across the floor with a pulling force (Fp) of 200 N which acts at an angle of 30 degrees above the horizontal. The coefficient of kinetic friction is 0.25. What is the acceleration of the box? (Use g = 9.8 m/s2. Hint: the normal force is not equal to the weight)

Answer 1

Answer:

The acceleration of the box is 2.05 m/s²

Explanation:

The given parameters of the motion of the box are;

The mass of the box, m = 50 kg

The pulling force, $F_p$ acting on the box = 200 N

The angle at which the force acts, θ = 30° above the horizontal

The coefficient of kinetic friction, $\mu_k$ = 0.25

The normal reaction from the box resting on a flat surface, N = The weight of the box, W - The vertical component of the pulling force, $F_{py}$

N = W -  $F_{py}$ = m·g - $F_p$ × sin(θ)

Where;

g = The acceleration due to gravity = 9.8 m/s²

∴ N = W  - $F_{py}$ = m·g - $F_p$ × sin(θ) = 50 kg × 9.8 m/s² - 200 N × sin(30°)

∴ N = 490 N - 200 N × 0.5 = 390 N

The normal reaction, N = 390 N

The force of friction, $F_f$ = The coefficient of kinetic friction, $\mu_k$ × The normal reaction, N

∴ $F_f$ = $\mu_k$ × N = 0.25 × 390 N = 97.5 N

The net force, $F_{NET}$, acting on the block = The pulling force, $F_p$ - The friction force, $F_f$

∴  $F_{NET}$ = $F_p$ - $F_f$ = 200 N - 97.5 N = 102.5 N

$F_{NET}$ = 102.5 N

According to Newton's second law of motion on the net force acting on an object, we have;

$F_{NET}$ = m × a

Where;

a = The acceleration of the box

∴ a = $F_{NET}$/m = 102.5 N/(50 kg) = 2.05 m/s²

The acceleration of the box = a = 2.05 m/s².