A 1 kg object starts 3 meters from the ground on a 24 degree incline with a friction coefficient of 0.12. When it hits the ground, it collides and sticks to a 8 kg object. The coefficient friction on the horizontal surface is 0.24 How far do the objects travel away from the incline?
When calculating the frictional work along the incline plane would this be the formula W= -0.12(1)(9.8)(3cos24)
You got the distance travelled wrong and the friction force wrong.
In traveling from 3 meters ABOVE the ground on a 24 degree incline, the distance travelled is 3/sin 24 = 7.38 meters. The opposing friction force is
M g cos24*0.12 = 1.07 N. Available kinetic energy at the bottom is
M g*3.0 m - 1.07N*7.38 m = 21.5 J
Check my numbers
Yes, you are correct! To calculate the frictional work along the incline plane, you can use the formula:
W = -μNds
where:
μ is the coefficient of friction (in this case, 0.12),
N is the normal force acting on the object (equal to mg, where m is the mass of the object and g is the acceleration due to gravity),
ds is the displacement along the incline.
In this case, the mass of the object is 1 kg, so the normal force N = mg = (1 kg)(9.8 m/s^2) = 9.8 N. The displacement along the incline is given as 3 m, and the angle of the incline is 24 degrees.
To calculate the horizontal component of the displacement, you can use the formula:
dx = ds * cos(θ)
where θ is the angle of the incline. So, dx = 3 m * cos(24 degrees).
Substituting the values into the formula, we can calculate the frictional work:
W = -0.12 * (9.8 N) * (3 m * cos(24 degrees))
After calculating the frictional work, you can proceed to solve the rest of the problem to determine how far the objects travel away from the incline.