Incline Plane Problem given mass, friction, 2 velocities

[TehDarkArchon]

Homework Statement

A solid mass weighing 50 kg slides down an incline plane with the coefficient of kinetic friction being 0.05. The velocity is measured to be 2.00 m/s and 5.00 m/s at two points 15 m apart. What are the A) Magnitude and direction of all forces action on the solid mass B) Acceleration of the mass and C) Angle the plane is inclined at?

Homework Equations

Fg = mg
Fn = mgcos(theta)
fs = (uk x Fn)
mgsin(theta)

F = ma

The Attempt at a Solution

So far I have the force of gravity (Fg) being mg = (50kg)(-9.81 m/s^2) but Fn = mgcos(theta), and since this is not given how would I go about calculating this? Also for part B, I was under the impression that since there is kinetic friction, using the kinematic equations would not work in this situation. Any help is much appreciated!

 

[tiny-tim]

Welcome to PF!

Hi TehDarkArchon! Welcome to PF!

(have a mu: µ anad a theta: θ and try using the X² tag just above the Reply box )

… for part B, I was under the impression that since there is kinetic friction, using the kinematic equations would not work in this situation.

All the forces are constant (including the friction), so the acceleration is constant, and the standard constant acceleration equations apply.

Use one of them to answer B first, then call the angle θ and answer A and C …

what do you get? ​

 

[TehDarkArchon]

Thank you very much for the prompt response, and sorry about not taking the time to get a good understanding of the formatting before posting =\ For B I used the equation v_f² = v_o² + 2ad, which in my case is rearranged to get (25 m/s - 4 m/s)/(2 x 15 m) giving me 0.700 m/s². I am confused about the angle though...should I use another kinematic equation and solve for time and then split the plane into horizontal and verticle components? It seems pretty complicated that way..

 

[tiny-tim]

Hi TehDarkArchon!

(just got up :zzz: …)

For B I used the equation v_f² = v_o² + 2ad, which in my case is rearranged to get (25 m/s - 4 m/s)/(2 x 15 m) giving me 0.700 m/s².

Yup!

Im confused about the angle though...should
I use another kinematic equation and solve for time and then split the plane into horizontal and verticle components? It seems pretty complicated that way..

Forget about horizontal and vertical …

in problems like this, use along-the-slope and perpendicular-to-the-slope instead!

There's zero acceleration perpendicular-to-the-slope, so that tells you what N is (as a function of θ), that gives you the friction force, and then you use F_total = ma along-the-slope, and solve for θ so that a = 0.7.

 

[rwisz]

I would approach this problem by first identifying the known and unknown variables and then using the appropriate equations to solve for the unknowns.

For the known variables, we have the mass (m = 50 kg), the coefficient of kinetic friction (uk = 0.05), and the velocities at two points (v1 = 2.00 m/s and v2 = 5.00 m/s).

To find the magnitude and direction of all forces acting on the solid mass, we can use Newton's second law (F = ma). Since the mass is moving with a constant velocity, we know that the net force acting on it is equal to zero. Therefore, the forces acting on the mass must be balanced, and we can set up the following equation:

Fg + fs = m*a

where Fg is the force of gravity, fs is the force of kinetic friction, and a is the acceleration of the mass.

To find the force of gravity, we can use Fg = mg, where g is the acceleration due to gravity (9.81 m/s^2). This gives us Fg = (50 kg)(9.81 m/s^2) = 490.5 N.

To find the force of kinetic friction, we can use the equation fs = uk*Fn. Fn represents the normal force, which is the force exerted by the incline plane on the mass. Since the mass is sliding down the incline, the normal force will be perpendicular to the incline and equal to the component of the force of gravity in that direction. Therefore, we can use Fn = mgcos(theta), where theta is the angle of the incline plane.

To find theta, we can use the trigonometric relationship between the angle of the incline and the given velocities. We know that the incline is 15 m long and the mass travels with a velocity of 2.00 m/s and 5.00 m/s at two points on the incline. Therefore, we can set up the following equation:

tan(theta) = (v2-v1)/15

Solving for theta, we get theta = 18.4 degrees.

Now, we can use this value to find the normal force, which is equal to (50 kg)(9.81 m/s^2)cos(18.4) = 457.2 N.

Finally,