# Solving Friction Problem: Mass 5kg Trolley, 25° Slope, Coefficient 0.4

• MHB
• Shah 72
In summary, the trolley, with a mass of 5 kg, is rolling up a rough slope at an angle of 25 degrees to the horizontal. The coefficient of friction between the trolley and the slope is 0.4. It passes a point A with a speed of 12m/s and, using the equations for motion, we can calculate its speed when it passes A on its way back down the slope to be approximately 9.17 m/s. The trolley's acceleration is -7.85 m/s^2 when going uphill and its acceleration is 7.85 m/s^2 when going downhill.
Shah 72
MHB
A trolley of mass 5 kg is rolling up a rough slope, which is at an angle of 25 degree to the horizontal. The coefficient of friction between the trolley and the slope is 0.4. It passes a point A with speed 12m/s. Find its speed when it passes A on its way back down the slope.
So I did F=m×a
-0.4×50cos25- 50sin25= 5a
a=-7.85m/s^2.
I don't know how to calculate after this. Pls help

let uphill be positive …

initial velocity > 0, acceleration < 0

$a = -g(\sin{\theta} + \mu\cos{\theta})$

$\Delta x = \dfrac{0^2-12^2}{2a} = 9.17 \, m$let downhill be positive …

initial velocity = 0, final velocity > 0, acceleration > 0, delta x > 0

$a = g(\sin{\theta} - \mu \cos{\theta})$

$v_f = \sqrt{ 0^2 + 2a \Delta x}$

skeeter said:
let uphill be positive …

initial velocity > 0, acceleration < 0

$a = -g(\sin{\theta} + \mu\cos{\theta})$

$\Delta x = \dfrac{0^2-12^2}{2a} = 9.17 \, m$let downhill be positive …

initial velocity = 0, final velocity > 0, acceleration > 0, delta x > 0

$a = g(\sin{\theta} - \mu \cos{\theta})$

$v_f = \sqrt{ 0^2 + 2a \Delta x}$
Thank you so much!

## 1. What is friction and how does it affect the movement of objects?

Friction is a force that opposes the motion of an object when it comes into contact with another surface. In this problem, friction is acting on the trolley as it moves down the slope, slowing down its movement.

## 2. How is the coefficient of friction determined?

The coefficient of friction is determined by dividing the force of friction by the normal force, which is the force perpendicular to the surface that the object is resting on. In this problem, the coefficient of friction is given as 0.4, meaning that the force of friction is 40% of the normal force.

## 3. What is the relationship between mass and friction?

The mass of an object does not directly affect the force of friction. However, a heavier object may experience more friction due to its weight and the normal force it exerts on the surface. In this problem, the mass of the trolley is 5kg, which will affect the normal force and therefore the force of friction.

## 4. How does the slope angle affect the force of friction?

The steeper the slope, the greater the force of friction acting on the object. This is because as the slope angle increases, the normal force also increases, resulting in a higher force of friction. In this problem, the slope angle is 25°, which will have a greater effect on the force of friction compared to a smaller slope angle.

## 5. How can friction be reduced or eliminated?

Friction can be reduced or eliminated by using lubricants, such as oil or grease, between surfaces. Additionally, reducing the contact area between surfaces or using smoother materials can also decrease friction. In this problem, the coefficient of friction can be reduced by using a lubricant on the surface of the slope or the trolley's wheels.

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