Question

A block of mass 10.0 kg is pulled to the right along a rough horizontal surface with a constant horizontal force of 20.0 N. The friction force is known to be 5.00 N. What is the magnitude of the acceleration of the block

Answer 1

The magnitude of the acceleration of the block is 1.5 m/s² to the right.

Step-by-step explanation:

The magnitude of the acceleration of the block can be determined using Newton's second law of motion. The net force acting on the block is the difference between the applied force and the frictional force. Thus, the net force is 20.0 N - 5.00 N = 15.0 N. According to Newton's second law, the net force is equal to the mass of the block multiplied by the acceleration. Rearranging the equation, we can solve for the acceleration:

15.0 N = 10.0 kg x acceleration

Solving for acceleration:

acceleration = 15.0 N / 10.0 kg = 1.5 m/s² to the right

Answer 2

the magnitude of acceleration will be 1.50m/s^2

Step-by-step explanation:

To calculate your acceleration, you can use your formula that states that the net force on an object is equal to the mass of the object multiplied by the acceleration of the object. Fnet=ma

if you draw out this situation and label the forces you will have your vector towards the right with a magnitude of 20.0N and then your friction vector will be pointing to the left (in other words, in the negative direction) (opposing the direction of movement) with a magnitude of 5.00N, with the 10.0 kg box in the middle.

The net force will be calculated using F1+F2=Fnet where your F1=20.0N and F2= -5.00N (since it is towards the negative direction).

you will find that Fnet=15.0N

With that, plug in the values you know to calculate the acceleration of the block:

Fnet=ma

(15.0N)=(10.0kg)a from her you can divide both sides by 10 to isolate a:

1.50=a (and now make sure to label the units of your answer)

a=1.50m/s^2 (which is the typical unit for acceleration)