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marysya [2.9K]
3 years ago
10

A sled is on an icy (frictionless) slope that is 30° above the horizontal. When a 40-N force, parallel to the incline and direct

ed up the incline, is applied to the sled, the acceleration of the sled is 2.0 m/s 2, down the incline.. The mass of the sled is:
Physics
1 answer:
son4ous [18]3 years ago
4 0

Answer: 5.8kg

Explanation:

F - mg sinΦ = ma

Given that

M=?

g= 9.8

Φ= 30

a= 2

F= 40 then

40 - 9.8 * m * sin 30 = 2 * m

40 = 2 * m + 9.8 * m * sin 30

40 = (2 + 9.8 * sin 30) m

m = 40 / (2 + 9.8 * 0.5)

m = 40 / (2 + 4.9)

m = 40 / 6.9

m = 5.797kg

m = 5.8kg

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Answer:

a)  W=0, b) Work is negative, c) work is positive and scientific energy variation is positive, d)     the variation of the potential enrgy is negative,

e) total work is positive

Explanation:

Work in physics is defined by the scalar scalar product of force by displacement

          W = F. dx

The bold are vectors; this can be written in the form of the mules of the quantities

          W = F dx cos θ

where θ is the angle between force and displacement.

a) The normal force is perpendicular to the inclined plane which is perpendicular to the displacement, therefore the angle is

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b) The friction force opposes the displacement whereby the angle is

       θ = 180      cos 190 = -1

        W = - fr d

Work is negative

c) To calculate the change in kinetic energy we use that the work is equal to the variation of the kinetic energy

            m g sin θ  L = ΔK

this magnitude is positive since the angle is zero cos 0 = 1

how the system starts from rest ΔK = Kf -K₀=  + Kf -0

work is positive and scientific energy variation is positive

d) change in potential energy

               The potential energy is is ΔU = Uf -U₀

we fix the reference system in the bases of the plane so Uf = 0

               ΔU = -U₀

         the variation of the potential enrgy is negative

e) The total work is formed by the work of the weight component, the work of the friction force

              W_Total = W_weight - W_roce

as the body moves down

              W_Total> 0

Therefore the total work is positive

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A man does 4,780 J of work in the process of pushing his 2.70 103 kg truck from rest to a speed of v, over a distance of 25.5 m.
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Answer:

(A) Velocity will be 1.88 m/sec

(b) Force will be 187.45 N

Explanation:

We have given work done = 4780 j

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(B) We know that work done is given by

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