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? .

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

Answer Link

johanrusli johanrusli · Answer 2 · 2019-12-13T09:01:16+01:00

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

[tex]\texttt{ }[/tex]

Further explanation

Let's recall Kinetic Energy Formula as follows:

[tex]Ek = \frac{1}{2}mv^2[/tex]

Ek = Kinetic Energy ( Joule )

m = mass of the object ( kg )

v = speed of the object ( m/s )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

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:

[tex]W_w = w \times h[/tex]

[tex]W_w = w \times d \times \sin \theta[/tex]

[tex]W_w = 2.0 \times 1.8 \times \sin 30^o[/tex]

[tex]W_w = 2.0 \times 1.8 \times \frac{1}{2}[/tex]

[tex]W_w = 1.8 \texttt{ Joule}[/tex]

[tex]\texttt{ }[/tex]

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

[tex]W_n = N \times d \times \cos \theta[/tex]

[tex]W_n = 1.7 \times 1.8 \times \cos 90^o[/tex]

[tex]W_n = 0 \texttt{ Joule}[/tex]

[tex]\texttt{ }[/tex]

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

[tex]W_f = -f \times d[/tex]

[tex]W_f = -\mu_k \times N \times d[/tex]

[tex]W_f = -0.30 \times 1.7 \times 1.8[/tex]

[tex]W_f = -0.918 \texttt{ Joule}[/tex]

[tex]\texttt{ }[/tex]

Learn more

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

[tex]\texttt{ }[/tex]

Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

[tex]\texttt{ }[/tex]

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