Question

A box of weight w=2.0N accelerates down a rough plane that is inclined at an angle ϕ=30∘ above the horizontal, as shown (Figure 6). The normal force acting on the box has a magnitude n=1.7N, the coefficient of kinetic friction between the box and the plane is μk=0.30, and the displacement d⃗ of the box is 1.8 m down the inclined plane. What is the work Ww done on the box by the weight of the box?

Answer 1

Answer:

The work done is 0.882 Joules.

Explanation:

To calculate the work, we need to find all forces that are involved in the movement.

As you can analyse, the body is the force that make the box to move, and the friction force is opposite to it, we cannot forget about the friction. So, we have to calculate the resultant force to this context.

$F_{R}=W_{x}-F_{k}$

So, to find $W_{x}$ we use: $W_{x}=W.sen\phi$; where $W=2N;\phi = 30\°$

$W_{x}=2N.sin30\°=1\frac{1}{2}N=1N$

The friction force would be:

$F_{k}=\mu_{k}N=(0.30)(1.7N)=0.51$

Then, the resultant force is:

$F_{R}=W_{x}-F_{k}$

$F_{R}=1N-0.51N=0.49N$

Now, we calculate the work: $W=F_{R} d$

$W=(0.49N)(1.8m)=0.882J$

Therefore, the work done is 0.882 Joules.

Answer 2

The work done on the box by the weight of the box is 1.8 Joule

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Further explanation

Let's recall Kinetic Energy and Work Formula as follows:

$Ek = \frac{1}{2}mv^2$

$W = Fd$

where:

Ek = Kinetic Energy ( Joule )

W = Work ( Joule )

m = mass of the object ( kg )

v = speed of the object ( m/s )

F = magnitude of force ( N )

d = displacement ( m )

Let us now tackle the problem !

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Given:

weight of the box = w = 2.0 N

angle of inclined plane = θ = 30°

normal force = N = 1.7 N

coefficient of kinetic friction = μ = 0.30

displacement of the box = d = 1.8 m

Asked:

work done by weight = W_w = ?

work done by normal force = W_n = ?

work done by the force of kinetic friction = W_fk = ?

Solution:

We could calculate work done by weight as follows:

$W_w = w \times h$

$W_w = w \times d \times \sin \theta$

$W_w = 2.0 \times 1.8 \times \sin 30^o$

$W_w = 2.0 \times 1.8 \times \frac{1}{2}$

$\boxed{W_w = 1.8 \texttt{ Joule}}$

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Because the direction of displacement is perpendicular to the direction of normal force , then we could calculate the work done by normal as follows:

$W_n = N \times d \times \cos \theta$

$W_n = 1.7 \times 1.8 \times \cos 90^o$

$W_n = 0 \texttt{ Joule}$

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Finally, we could calculate the work done by friction as follows:

$W_f = -f \times d$

$W_f = -\mu_k \times N \times d$

$W_f = -0.30 \times 1.7 \times 1.8$

$W_f = -0.918 \texttt{ Joule}$

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Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441
• Newton's Law of Motion: brainly.com/question/10431582
• Example of Newton's Law: brainly.com/question/498822

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics