How do you calculate the work done when lifting an object?

If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric

poizon [28]

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

3 0

2 years ago

2

timama [110]

Answer:

A

Explanation:

The acceleration of an object is directly proportional to its net force.

$a = \frac{f}{m}$

4 0

3 years ago

Read 2 more answers

Which chemical can be toxic to the cells of it’s not removed?

masya89 [10]

Hey There,

Question: Which chemical can be toxic to the cells of it’s not removed?

Answer: D. Carbon Dioxide

If This Helps May I Have Brainliest?

5 0

3 years ago

Read 2 more answers

A locomotive accelerates a 25-car train along a level track. Every car has a mass of 7.7 ✕ 104 kg and is subject to a friction f

MaRussiya [10]

To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.

In this way,

The total mass of the cars would be,

$m_T = 25(7.7*10^4)Kg$

$m_T = 1.925*10^6Kg$

Therefore the friction force at 29Km / h would be,

$f=250v$

$f= 250*29Km/h$

$f = 250*29*(\frac{1000m}{1km})(\frac{1h}{3600s})$

$f = 2013.889N$

In this way the tension exerts between first car and locomotive is,

$T=m_Ta+f$

$T=(1.925*10^6)(0.2)+2013.889$

$T= 3.8701*10^5N$

Therefore the tension in the coupling between the car and the locomotive is $3.87*10^5N$

6 0

3 years ago

If μs is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. calculate this

fiasKO [112]

weight = mg acts downwards 
normal force = N acts upwards.
and force F acts at an angle θ below the horizontal.
(Let us assume that the woman pushes from the left, so F is acted towards the right, which is below the horizontal)
so that, Frictional force, f=us*N acts towards the left

Now we balance the forces along x and y directions:
y direction: N = mg + F sinΘ
x direction: us * N = F cosΘ

We let the value of µs be equal to a value such that any F will not be able to move the crate. Then, if we increase F by an amount F', then the force pushing the crate towards the right also increases by F' cosΘ. Additionally, the frictional force f must raise by exactly this amount.
Since f can’t exceed us*N, so the normal force must increase by F' cosΘ/us.
Also, from the y direction equation, the normal force exceeds by F' sin Θ.

These two values must be the same, therefore:
us = cot θ

4 0

3 years ago