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IceJOKER [234]
3 years ago
8

An upward force is applied to a 6.0–kilogram box. This force displaces the box upward by 10.00 meters. What is the work done by

the force on the box?
a. 6.0 × 101 joules
b. -6.0 × 101 joules
c. 5.9 × 102 joules
d. -5.9 × 102 joules
Physics
1 answer:
devlian [24]3 years ago
3 0

Answer:

W ≈ 5.9 × 10^{2} J

Hence, option (c) is correct.

Given:

mass of box = 6 kg

Displacement of box = 10 m

To find:

Work done by the force = ?

Formula used:

Work done is equal to change is potential energy,

W = m g h

Where, W = Work done

m = mass of the box

h = displacement by the force

Solution:

Work done is equal to change is potential energy,

W = m g h

Where, W = Work done

m = mass of the box

h = displacement by the force

W = 6 × 9.8 ×10

W = 588 Joule

Thus, W ≈ 5.9 × 10^{2} J

Hence, option (c) is correct.

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b) 0 kg m/s

c) +3 m/s

d) 60 N

Explanation:

a)

The momentum of an object is a vector quantity given by:

p=mv

where

m is the mass of the object

v is the velocity of the object

In this problem, we have a system of two people, so the total momentum will be the sum of the individual momenta of the two people:

p=p_1 + p_2

Which can be rewritten as

p=m_1 u_1 + m_2 u_2

where m_1,m_2 are the masses of the two people and u_1,u_2 their initial velocities.

However, the two people are initially at rest, so

u_1 = 0\\u_2 = 0

Therefore the total momentum is

p=0+0=0

b)

The principle of conservation of momentum states that when there are no external forces acting on a system, the total momentum of the system is conserved, so we can write:

p_i = p_f

where

p_i is the total momentum of the system before

p_f is the total momentum of the system after

In this problem,

p_i = 0

As we calculated in part a: this is because the total momentum of the two people before they push off each other is zero.

Therefore, according to the law of conservation of momentum,

p_f = p_i = 0

So the total momentum is zero also after they push off each other.

c)

The total momentum of the girl and the boy after they push off each other can be written as:

p_f = m_1 v_1 + m_2 v_2 (1)

where:

m_1 = 30 kg is the mass of the girl

v_1 = -5 m/s is her velocity (she moves backward, so the negative sign)

m_2 = 50 kg is the mass of the boy

v_2 is the velocity of the boy

As calculated in part b), we also know that the total momentum of the girl and the boy is

p_f = 0 (2)

By combining eq(1) and eq(2) we get

0=m_1 v_1 + m_2 v_2

And solving for v2 we find the velocity of the boy:

v_2=-\frac{m_1 v_1}{m_2}=-\frac{(30)(-5)}{50}=+3 m/s

and the positive sign means he is moving forward.

d)

We can solve this part by applying the impulse theorem, which states that the change in momentum of an object is equal to the product between the force applied on it and the duration of the collision:

\Delta p = F\Delta t

where

\Delta p is the change in momentum

F is the force

\Delta t is the time during which the force is applied

In this problem:

\Delta t = 2.5 s

For the boy, the change in momentum is:

\Delta p = m_2 (v_2 - u_2)

And since

m_2 = 50 kg\\u_2 = 0 m/s\\v_2 = 3 m/s

We have

\Delta p = (50)(3-0)=150 kg m/s

So, the force exerted between the boy and the girl is:

F=\frac{\Delta p}{\Delta t}=\frac{150}{2.5}=60 N

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