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irina [24]
3 years ago
15

In pushing a heavy box across the floor, is the force you need to apply to start the box moving greater than, less than, or the

same as the force needed to keep the box moving? On what are you basing your choice?
How do you think the force of friction is related to the weight of the box? Explain.
Physics
1 answer:
Veseljchak [2.6K]3 years ago
3 0

Answer:

Explanation:

Force needed to apply start the box is greater than the force needed to keep it moving because static friction is greater than the kinetic friction .

A  threshold force is needed to move the box and when box started to move kinetic friction comes into play.      

Friction force is directly related to the weight of the box as the friction force is

coefficient of friction time Normal reaction .

And Normal reaction is equal to the weight of box if no force is applied.

f_r=\mu N

N=mg

You might be interested in
The question is in the picture
Sedbober [7]

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

v_0-gt = 0.

where v_0 is the initial velocity.

Solving for t we get

t = \dfrac{v_0}{g}

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time t_0 since it had reached the highest point, the ball has traveled downwards and the velocity v_f it has gained is

v_f = gt_0,

and we are told that this is twice the initial velocity v_0; therefore,

v_f = 2v_0  = gt_0

which gives

t_0 = \dfrac{2v_0}{g}.

Thus, the total time taken to reach velocity 2v_0 is

t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}

t_{tot} = \dfrac{3v_0}{g}.

This t_{tot}, we are told, is 36 seconds; therefore,

36= \dfrac{3v_0}{g},

and solving for v_0 we get:

v_0 = \dfrac{36g}{3}

v_0 = \dfrac{36s(10m/s^2)}{3}

\boxed{v_0 = 120m/s}

which from the options given is choice e.

7 0
3 years ago
QUESTION 1
Molodets [167]
Question 1: C Question 2: B, Hope this Helps!
3 0
2 years ago
Puck B has twice the mass of puck A. Starting from rest, both pucks are pulled the same distance across frictionless ice by stri
gogolik [260]

Answer:

(a) 1 : 2

(b) same

Explanation:

Let the mass of puck A is m and the mass of puck B is 2 m.

initial speed for both the pucks is same as u and the distance is same for both is s.

let the tension is T for same.

The kinetic energy is given by

K = 0.5 mv^2

(a) As the speed is same, so the kinetic energy depends on the mass.

So, kinetic energy of A : Kinetic energy of B = m : 2m  = 1 : 2

(b) A the distance s same so the final velocities are also same.

8 0
2 years ago
A crate lies on a plane tilted at an angle θ = 22.5 ∘ to the horizontal, with μk = 0.19.
ValentinkaMS [17]

A) 2.03 m/s^2

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

mg sin \theta - \mu_k R = ma (1)

where

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

\theta=22.5^{\circ}

\mu_k = 0.19 is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

R-mg cos \theta =0 (2)

From (2) we find

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_k mg cos \theta = ma

Solving for a,

a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2

B) 5.94 m/s

We can solve this part by using the suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = ?

u = 0 (it starts from rest)

a=2.03 m/s^2

s = 8.70 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s

6 0
3 years ago
Which best describes the transition from gas to liquid? (2 points) Select one: a. Energy must be removed because particles in li
Alenkinab [10]

Which best describes the transition from gas to liquid?

gas is @ higher energy state than liq. so the transition must remove energy. so ans is  a. Energy must be removed because particles in liquid move more slowly.

8 0
3 years ago
Read 2 more answers
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