What is the force of friction, in n, acting on a 25 kg sled being pulled with a rope with a force of tension of 50 n when moving a constant speed of 4 m/s?

the answer is 50 n

but how do i get the answer?

Frictional force is 50 N.

Explanation:

Let the frictional force acting on the sled be 'f'.

Given:

Mass of the sled is,

Force of pull is,

The sled moves with a constant speed of 4 m/s.

Now, we know that, acceleration of an object is equal to the rate at which its velocity changes.

So, if the velocity doesn't change, then the acceleration of the object will be zero. Here, the sled is moving at a constant speed without changing its direction. So, velocity here is a constant. Therefore, acceleration of the sled is zero as the change in velocity is zero.

Now, forces acting on the sled are force 'F' and frictional force 'f' in the opposite direction.

Therefore, net force acting on the sled is given as:

Now, from Newton's second law, we have

Net force = mass acceleration.

Therefore, the frictional force acting on the sled will be equal to the applied force but in the opposite direction so that the net force becomes zero.

Therefore, frictional force = 50 N

hi,

what occurs during a solar eclipse?

there are small tides across earth.

the moon is between earth and the sun.

p.s. im not sure if the first answer is correct because during a solar ecliphse there are spring tides and i dont know if they are the same.

there are no atoms