A student lifts a set of books off a table and places them in the upper shelf of a book case which is 2 meters above the table.
2 answers:
The work done is the product between the intensity of the force applied F, the amount of the displacement d of the book and the cosine of the angle 
between the direction of the force and the direction of the displacement:
In our problem, the student is lifting the book, so he is applying a force directed upward, and the book is moving upward, so F and d are parallel and therefore the angle is zero, so
Therefore, the work done is
In our problem, the student is lifting the book, so he is applying a force directed upward, and the book is moving upward, so F and d are parallel and therefore the angle is zero, so
Therefore, the work done is
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As per given conditions there are two directions along which forces are acting
1. Net force along left direction is given as
2. Net force towards right direction is given as
now since the two forces here in opposite direction so here we will have net force given as
so here net forces must be 440 N towards right
Answer:
A) The free body diagrams for both the load of bricks and the counterweight are attached.
B) a = 2.96 m/s²
Explanation:
A)
The free body diagrams for both the load of bricks and the counterweight are attached.
B)
The acceleration of upward acceleration of the load of bricks is given by the following formula:
a = g(m₁ - m₂)/(m₁ + m₂)
where,
a = upward acceleration of load of bricks = ?
g = 9.8 m/s²
m₁ = heavier mass = mass of counterweight = 28 kg
m₂ = lighter mass = mass of load of bricks = 15 kg
Therefore, using these values in equation, we get:
a = (9.8 m/s²)(28 kg - 15 kg)/(28 kg + 15 kg)
<u>a = 2.96 m/s²</u>
