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 . .
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What is the work Ww done on the box by the weight of the box? . Express your brianowens.tvs 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 !
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
work done by weight = W_w = ?
work done by normal force = W_n = ?
work done by the force of kinetic friction = W_f = ?
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
Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant