A rope is used to pull a 3.57 kg block at constant speed 4.06 m along a horizontal floor. The force on the block from the rope is 7.68 N and directed 15.0° above the horizontal. What are (a) the work done by the rope’s force, (b) the increase in thermal energy of the block–floor system, and (c) the coefficient of kinetic friction between the block and floor?

μ= 0.22

Explanation:

Work done by the rope's force= F×s×cos θ= 7.68×4.06× cos 15.0°= 30.1J

b) work done by rope= increase in thermal energy= 30.1J

c) From

F cos θ = μ(mg- Fsinθ)

μ= F cos θ/mg- Fsinθ

μ= 7.68 cos 15.0°/(3.57×10)-(7.68 sin 15.0°)

μ= 0.22

A) W = 30.12J

B) Δthermal = 30.12 J

Explanation:

We are given that ;

Mass; m = 3.57 kg

Force on block; F = 7.68N

θ = 15°

d = 4.06m

The work done by the rope forces will be;

W = Fd cosθ = 7.68 x 4.06 x cos 15 = 30.12J

B) The block moves with constant speed and so the pulling force will be equal to the friction force.

Thus;

the increase in thermal energy of the block–floor system will be equal to work done; thus;

Δthermal = 30.12 J

C) To calculate this, resolving forces, we obtain;

Fcos θ = μ(mg - F sin θ)

Where μ is coefficient of kinetic friction.

Thus, making μ the subject, we have;

Fcosθ/(mg - F sin θ) = μ

Plugging in relevant values to obtain;

μ = 7.68cos15/(3.57x9.8 - 7.68 sin 15) = 7.4183/(34.986 - 1.9877) = 0.225

there are no components of a physical goal mentioned on the list of choices that you've provided.

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j