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DanielleElmas [232]
3 years ago
7

a 1811kg car speeds up from rest when a traffic light turns green until it reaches a speed of 23m/s. how much work was done on t

he car by friction?
Physics
1 answer:
sattari [20]3 years ago
3 0

Answer:

The work done by the frictional force, W = 612565.32 J

Explanation:

Given data,

The mass of the car, m = 1811 kg

The initial velocity of the car, u = 0 m/s

The final velocity of the car, v = 23 m/s

Let the time period of the car be, t = 3 s

The acceleration of the car,

                              a = (v-u) /t

                                 = (23 - 0)/ 3

                                = 7.67 m/s²

The normal force acting on the car,

                         F = mg

                             = 1811 kg x 9.8 m/s²

                              = 17747.8 N

The displacement of the car,

                             s = ut + ½ at²

                                 = 0 + ½ x 7.67 x 3²

                                 = 54.52 m

The work done by the frictional force,

                              W = F · S

                              W = 17747.8 N x 54.52 m

                                   = 612565.32 J

Hence, the work done by the frictional force, W = 612565.32 J

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A small box is held in place against a rough vertical wall by someone pushing on it with a force directed upward at 27 ∘ above t
ycow [4]

Answer:

Explanation:

Applied force, F = 18 N

Coefficient of static friction, μs = 0.4

Coefficient of kinetic friction, μs = 0.3

θ = 27°

Let N be the normal reaction of the wall acting on the block and m be the mass of block.

Resolve the components of force F.

As the block is in the horizontal equilibrium, so

F Cos 27° = N

N = 18 Cos 27° = 16.04 N

As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .

The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N   .... (1)

The vertically downward force acting on the block is mg - F Sin 27°

                                                      = mg - 18 Sin 27° = mg - 8.172    ... (2)

Now by equating the forces from equation (1) and (2), we get

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m = 1.49 kg

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