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3 years ago

Will the velocity of a box change as it moves across the surface with no friction?Explain

1 answer:

3 0

The velocity will remain unchanged (Newton’s 1st Law) unless a force acts on it. If no friction force, then the velocity remains constant.

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Which poison was used in the Jonestown massacre?

olga nikolaevna [1]

That would be Cyanide.

Hope this helps! (:

3 0

3 years ago

Read 2 more answers

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

Mariana [72]

a) $F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

Here the crate is moving at constant velocity, so no acceleration:

a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

- Vertical direction: the equation of the forces is

$R-Fsin \theta - mg = 0$ (1)

where

R is the normal reaction of the floor (upward)

$F sin \theta$ is the component of the force F in the vertical direction (downward)

mg is the weight of the crate (downward)

- Horizontal direction: the equation of the forces is

$F cos \theta - \mu_k R = 0$ (2)

where

$F cos \theta$ is the horizontal component of the force F (forward)

$\mu_k R$ is the force of friction (backward)

From (1) we get

$R=Fsin \theta +mg$

And substituting into (2)

$F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

b) $\mu_s=cot(\theta)$

In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

$R-Fsin \theta - mg = 0$ (1)

$F cos \theta - \mu_s R = 0$ (2)

In this second case, we want to find the critical value of $\mu_s$ such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction, $F cos \theta$ .

Therefore, using the same procedure as before,

$R=Fsin \theta +mg$

$F cos \theta - \mu_s (Fsin \theta +mg) = 0$

And solving for $\mu_s$ ,

$F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}$

Now we analyze the expression that we found. We notice that if the force applied F is very large, $F sin \theta >> mg$ , therefore we can rewrite the expression as

$\mu_s \sim \frac{F cos \theta}{F sin \theta}\\\mu_s=cot(\theta)$

So, this is the critical value of the coefficient of static friction.

8 0

3 years ago

What is the pressure exerted by 0.801 mol of co2 in a 12 l container?

Svetlanka [38]

We assume that the gas is an ideal gas so we can use the relation PV=nRT. Assuming that the temperature of the system is at ambient temperature, T = 298 K. We can calculate as follows:

PV = nRT
P = nRT / V
P = (0.801 mol ) (0.08205 L-atm / mol-K) (298.15 K) / 12 L
P = 1.633 atm

8 0

3 years ago

Two uniform solid spheres, A and B have the same mass. The radius of sphere B is twice that of sphere A. The axis of rotation pa

Svetach [21]

Answer:

Sphere B has 4 times more inertia than sphere A.

Explanation:

Inertia on a Solid sphere is given by:

$I = 2/5*M*R^2$