Un pintor de 75.0 kg sube por una escalera de 2.75 m que está inclinada contra una pared vertical. La escalera forma un ángulo d
1 answer:
Answer:
Work done, W = 1786.17J
Explanation:
The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "
Mass of a painter, m = 75 kg
He climbs 2.75-m ladder that is leaning against a vertical wall.
The ladder makes an angle of 30 degrees with the wall.
We need to find the work done by the gravity on the painter.
The angle between the weight of the painter and the displacement is :
θ = 180 - 30
= 150°
The work done by the gravity is given by :
Hence, the required work done is 1786.17 J.
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Answer:
Explanation:
v = Velocity of Ferdinand = 5 m/s
= Angle of jump
T = Time taken = 0.6 s
R = Distance between lily pads = 2.4 m
Horizontal range is given by
The angle at which Ferdinand make each of his jumps is
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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