Abox 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 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? .

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

Let's recall Kinetic Energy Formula as follows:

Ek = Kinetic Energy ( Joule )

m = mass of the object ( kg )

v = speed of the object ( m/s )

Let us now tackle the problem !

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:

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

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

Grade: High School

Subject: Physics

Chapter: Dynamics

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

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.