Question

An 7.6 kg crate is pulled 4.8 m up a 30∘ incline by a rope angled 18 ∘ above the incline. The tension in the rope is 130 N and the crate's coefficient of kinetic friction on the incline is 0.22.1.) How much work is done by (a) tension, by (b) gravity, and by (c) the normal force?2. ) What is the increase in thermal energy of the crate and incline?

Answer 1

Answer:

Part a)

$W_T = 593.4 J$

Part b)

$W_{mg} = -179 J$

Part c)

$W_{N} = 0$

Part 2)

$W_{thermal} = 25.76 J$

Explanation:

Part a)

As we know that the rope makes an angle of 18 degree above the inclined plane

Here we know that the tension in the rope is T = 130 N

displacement of the block is

d = 4.8 m

now the work done by the tension force is given as

$W = T . d$

$W = T d cos\theta$

$W = (130)(4.8) cos18$

$W = 593.4 J$

Part b)

Work done due to gravity is given as

$W = -(F_g sin\phi)(d)$

here we know that the angle of inclined plane is

$\phi = 30 Degree$

so we will have

$W = -(mg sin\phi)(d)$

$W = -(7.6 \times 9.81) sin30 (4.8)$

$W = -179 J$

Part c)

As we know that normal force is always perpendicular to the displacement

so here we know that

$W = F . d cos 90$

$W = 0$

Part 2)

Increase in thermal energy of the crate is equal to the work done by frictional force

so we will have

$W = F_f d$

here we know that

$F_f = \mu F_n$

where normal force is given as

$F_n = mgcos\phi - T sin\theta$

$W = \mu (mg cos\phi - T sin\theta)d$

$W = 0.22(7.6 (9.81)cos30 - 130 sin18)(4.8)$

$W = 25.76 J$

Answer 2

The work done by tension in pulling the crate is 593.42 J.

The work done by gravity on the crate is 178.75 J.

The work done by normal force is 0 J.

The increase in the thermal  energy of the crate and incline is 25.7 J.

The given parameters;

• mass of the crate, m = 7.6 kg
• distance traveled by the crate, d = 4.8 m
• inclination of the plane, θ = 30⁰
• inclination of the rope, = 18⁰
• tension on the rope, T = 130 N

The work done by tension is calculated as follows;

$W= Tcos(\theta)\times d\\\\W= 130 cos(18) \times 4.8 = 593.42 \ J$

The work done by gravity is calculated as follows;

$W = F_g sin(\theta) \times d\\\\W = mgsin(\theta) \times d\\\\W = (7.6 \times 9.8)\times sin(30) \times 4.8 = 178.75 \ J$

The work done by normal force is calculated as follows;

$W = F cos(90) \times d\\\\W = 0$

The increase in the thermal  energy of the crate and incline is calculated as follows;

$W = F_k \times d\\\\W = \mu_k F_n \times d\\\\W = \mu_k(Wcos(\theta) - Tsin(\theta) \times d\\\\W = 0.22(7.6\times 9.8cos(30) - 130sin(18)) \times 4.8\\\\W = 25.7 \ J$

Learn more here:https://brainly.com/question/12797227