Answer :
Final answer:
The frictional force acting on a 12.0 kg box with a coefficient of kinetic friction of 0.510 is 26.95 N, and the acceleration of the box given an applied force of 235 N at an angle of 32° is 14.38 m/s².
Explanation:
To solve this problem, we need to apply Newton's second law of motion, which states that force equals mass times acceleration (F = ma), and consider the effects of frictional force which opposes the motion of objects sliding against each other.
Magnitude of Frictional Force (a)
The frictional force can be calculated using the following steps:
Calculate the normal force (N) acting on the box. N = m * g - F * sin(\theta), where m is the mass, g is the gravitational acceleration (9.8 m/s²), and \theta is the angle of applied force.
Use the coefficient of kinetic friction (\(\mu_k\)) to find the frictional force (f_k). f_k = \(\mu_k\) * N.
Magnitude of the Acceleration (b)
To find the acceleration:
- Calculate the net force acting on the box. Subtract the frictional force from the horizontal component of the applied force.
- Use Newton's second law (F_net = m * a) to solve for acceleration (a).
Let's calculate these step-by-step.
1. Normal force: N = (12.0 kg)(9.8 m/s²) - (235 N) sin(32.00°)
= 118.6 N - 124.1 N * 0.5299
= 118.6 N - 65.76 N
= 52.84 N
2. Frictional force: f_k = (0.510)(52.84 N) = 26.95 N
3. Net force: F_net = (235 N) cos(32.00°) - 26.95 N
= 199.48 N - 26.95 N
= 172.53 N
4. Acceleration: a = F_net / m
= 172.53 N / 12.0 kg
= 14.38 m/s²
