Question

A 7.0-N force parallel to an incline is applied to a 1.0-kg crate. The ramp is tilted at 20° and is frictionless.

(a) What is the acceleration of the crate?
(b) If all other conditions are the same but the ramp has a friction force of 1.9 N, what is the acceleration?

Answer 1

The acceleration of the crate on the frictionless ramp is 2.39 m/s², while the acceleration on the ramp with friction is 0.49 m/s².

Step-by-step explanation:

To find the acceleration of the crate, we first need to determine the force component parallel to the incline. This can be done using the formula F_parallel = F * sin(theta), where F is the applied force and theta is the angle of the ramp. Substituting the values, we have F_parallel = 7.0 N * sin(20°) = 2.39 N. The acceleration can then be found using the equation a = F_parallel / m, where m is the mass of the crate. Substituting the values, we have a = 2.39 N / 1.0 kg = 2.39 m/s².

If the ramp has a friction force of 1.9 N, it will oppose the motion of the crate. In this case, we need to subtract the friction force from the parallel force to find the effective force. The effective force, Feff, is given by Feff = F_parallel - friction force. Substituting the values, we have Feff = 2.39 N - 1.9 N = 0.49 N. The acceleration can then be found using the equation a = Feff / m. Substituting the values, we have a = 0.49 N / 1.0 kg = 0.49 m/s².

Answer 2

a) 0.1m/s^2 b) 0.12m/s^2

Step-by-step explanation:

a) Using the concept of friction to solve the question,

First we need to know all the forces acting on the body on the inclined plane

Forces acting parallel to the plane are the moving force (Fm), frictional force(Ff) and mgsintheta (resolving weight along the plane).

Force acting perpendicular to the plane are weight (mgcostheta) and the normal reaction.

Using the formula

Summation of forces along the plane = mass × acceleration

Fm+mgsintheta -Ff= ma...(1)

Fm= 7.0N

Mass= 1.0kg

Ff=0 (since the movement is frictionless)

Substituting the datas in (1)

7.0 + 1×10sin20-0 = 1.0a

7.0+3.42=1.0a

a = 1.0/10.42

a= 0.1m/s^2

b) Now that the fictional force is 1.9N, the acceleration will change. Using the same formula in (a)

7.0+1×10sin20-1.9 = 1.0a

7.0+3.42-1.9=1.0a

8.52 = 1.0a

a = 1.0/8.52

a=0.12m/d^2