Ask question

Ask question

All categories

1 month ago

11

A force of 150 N accelerates a 25 kg wooden chair across a wood floor at 4.3 m/s2 . How big is the frictional force on the block

? What is the coefficient of friction between the chair and floor?

1 answer:

Ostrovityanka [3.2K]1 month ago

4 0

Initially, we will determine the net force based on the provided details. According to Newton's second law, F(net) = ma: F(net) = 25 * 4.3 = 107.5. Next, we will find the frictional force, f, which counters the applied force, F(app), explaining why the net force is slightly reduced: f = F(net) - F(app) = 150 - 107.5 = 42.5 N. We can then compute the coefficient of friction, u, using the normal force, F(N): f = uF(n) --> u = f/F(N), yielding u = 42.5/[25(9.8)] = 0.17.

You might be interested in

A 3030 cmcm wrench is used to loosen a bolt with a force applied 0.30.3 mm from the bolt. It takes 6060 NN to loosen the bolt wh

Softa [3030]

Complete Question

A 30 cm cm wrench is employed to loosen a bolt, with force applied 0.30m from the bolt. It requires 60 N to loosen the bolt when force is applied perpendicular to the wrench. How much force would be necessary if the force were applied at a 30-degree angle from perpendicular?

Response:

The strength needed is $F_{\theta } = 69.28 \ N$

Clarification:

According to the problem, we know that

The length of the wrench is $L = 30 cm = \frac{30}{100} = 0.3 \ m$

The distance from the bolt is $d = 0.30 m$

The force necessary to loosen the bolt is $F = 60 N$

The angle of application is $\theta = 30 ^o$

Generally, the torque required for loosening the bolt is defined as

$\tau = F * d$

$\tau = 60 * 0.3$

$\tau = 18 Nm$

Now for the bolt to loosen at $\theta$ the torque at 90° must equal that at $\theta$

Thus, the torque at $\theta$ is represented mathematically as

$\tau = F_{\theta }d cos \theta$

substituting values gives us

$18 = F_{\theta } * 0.3 cos (30)$

$F_{\theta } = \frac{18}{0.3 cos (30)}$

$F_{\theta } = 69.28 \ N$

7 0

1 month ago

1. The gravitational pull of the sun on Earth keeps Earth orbiting around the sun. Which statement is correct about the force th

Sav [3153]

Answer:

1. The sun is moved away from the Earth by an equal force exerted by the Earth.

6 0

2 months ago

Read 2 more answers

Two basketball players are essentially equal in all respects. (They are the same height, they jump with the same initial velocit

kicyunya [3294]

Boris's reaction time is denoted as t(r), implying that he has not jumped prior to that moment. Therefore, H(b)(t) equals 0
. The vertical displacement is determined simply as
D(t) = H(a)(t)

7 0

2 months ago

A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn

Ostrovityanka [3204]

The alteration in potential energy is $\Delta PE = - 3.8*10^{-16} \ J$

In the query, it is stated that

The intensity of the uniform electric field equals $E = 950 \ N/C$

The distance the electron covers is $x = 2.50 \ m$

Typically, the force exerted on this electron is expressed mathematically as

$F = qE$

Where F signifies the force and q represents the charge of the electron, which is a fixed value of $q = 1.60*10^{-19} \ C$

Thus

$F = 950 * 1.60 **10^{-19}$

$F = 1.52 *10^{-16} \ N$

Generally, the work-energy theorem is mathematically framed as

$W = \Delta KE$

Where W denotes the work done on the electron by the electric field and $\Delta KE$ is the change in kinetic energy

Additionally, work done on the electron can also be described as

$W = F* x *cos( \theta )$

Where $\theta = 0 ^o$ assuming that the electron's movement aligns with the x-axis

So

$\Delta KE = F * x cos (0)$

Inserting values

$\Delta KE = 1.52 *10^{-16} * 2.50 cos (0)$

$\Delta KE = 3.8*10^{-16} J$

According to the conservation of energy

$\Delta PE = - \Delta KE$

Where $\Delta PE$ signifies the change in potential energy

Thus

$\Delta PE = - 3.8*10^{-16} \ J$

7 0

2 months ago