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. . . I got Part A, need help with B and C . . What is the work Ww done on the box by the weight of the box? . Express your answers in joules to two significant figures. . . W=1.8J . . B)What is the work Wn done on the box by the normal force? .

Answer 1

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

The work done on the box by the normal force is 0 Joule

The work done on the box by the force of kinetic friction is -0.918 Joule

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Further explanation

Let's recall Kinetic Energy Formula as follows:

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

Ek = Kinetic Energy ( Joule )

m = mass of the object ( kg )

v = speed of the object ( m/s )

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_f = ?

Solution:

We could calculate work done by weight as follows:

$W_w = w \times h$

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

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

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

$W_w = 1.8 \texttt{ Joule}$

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Because the direction of displacement is perpendicular to the direction of normal force , then:

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

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

$W_n = 0 \texttt{ Joule}$

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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

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• Example of Newton's Law: brainly.com/question/498822

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

Answer 2

Work done = Force(in direction of movement) x displacement

Force from kinetic friction = friction coefficient x normal force. In this case the friction is acting antiparallel to the direction of motion with a force of 0.3 x 1.7 = 0.51N, again W=Fs, so work done on the box by kinetic friction = 0.51 x 1.8 = 0.918J