A 0.50-kg block slides across a tabletop with an initial velocity of 20 cm/s and comes to rest in a distance of 70 cm. Find the
2 answers:
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The magnitude of the friction force is 25 N
Explanation:
To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:
- The horizontal component of the pulling force,
, where F = 50 N is the magnitude and
is the angle between the direction of the force and the horizontal; this force acts in the forward direction
- The force of friction,
, acting in the backward direction
According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:
where
m is the mass of the block
is the horizontal acceleration
However, the block is moving at constant speed, so the acceleration is zero:
So the equation becomes
(1)
The net force here is given by
(2)
And so, by combining (1) and (2), we find the magnitude of the friction force:
Learn more about force of friction:
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Answer: 117.60N
Explanation:
Weight is a force. Therefore, we can use the force formula to find weight.
W = weight
m = mass
g = acceleration due to gravity ()