A 25 kg box slides, from rest, down a 9-m-long incline making an angle 15 degrees with the horizontal. The speed of the box when it reaches the bottom of the incline is 2.40 m/s.
(a) What is the coefficient of kinetic friction between the box and the surface of the incline?
(b) How much work is done on the box by the force of friction?
(c) What is the change in the potential energy of the box?
(a)
Distance, S = 9m
initial velocity, u = 0 m s-2
final velocity, v = 2.4 m s-2
acceleration, a (assumed uniform)
= (v²-u²)/2S
= (2.4²-0²)/(2*9)
= 0.32 ms-2
Acceleration due to gravity, g
= 9.8 m s-2
component of g along incline
= g sin(θ)
= 9.8 sin(15°)
= 2.536 m s-2
Therefore reduced acceleration due to friction, af
= 2.536-0.32
= 2.216 ms-2
Frictional force, F
= m(af)
= 25 kg * 2.216 m s-2
= 55.411 N
Normal force acting on inclined plane, N
= m * g cos(θ)
= 25 * 9.8 * cos(15°)
= 236.6 N
Coefficient of kinetic friction, μ
= F/N = 0.234
(b) work done = force * distance
(c) Change in potential energy = mgh
where h is the difference in elevation (final - initial).
if the mass were suspended by water and it's specific gravity were less than one would it knot float? slip up!
To answer these questions, we need to use several key physics concepts such as kinetic energy, potential energy, work, and Newton's laws of motion. Let's go step by step to find the answers.
(a) To find the coefficient of kinetic friction, we can use the formula for the net force along the incline:
Net force = (mass x acceleration) + (mass x g x sin(θ)) - (coefficient of kinetic friction x mass x g x cos(θ))
Given that the box slides down the incline from rest, we know that the initial velocity is zero (v0 = 0). Also, since the box reaches the bottom of the incline with a speed of 2.40 m/s (vf = 2.40 m/s), we can use the equation of motion vf^2 = v0^2 + 2ax to find the acceleration (a) along the incline.
a = (vf^2 - v0^2) / (2 x distance) = (2.40^2 - 0) / (2 x 9) = 0.64 m/s^2
Plugging in the values into the net force equation and solving for the coefficient of kinetic friction:
Net force = (25 kg x 0.64 m/s^2) + (25 kg x 9.8 m/s^2 x sin(15°)) - (coefficient of kinetic friction x 25 kg x 9.8 m/s^2 x cos(15°))
Remember to convert the angle from degrees to radians: sin(15°) = sin(15° x π/180) and cos(15°) = cos(15° x π/180).
Simplifying the equation and solving for the coefficient of kinetic friction:
Net force = 16 + (25 x 9.8 x sin(15°)) - (coefficient of kinetic friction x 25 x 9.8 x cos(15°))
Net force = 16 + (25 x 9.8 x 0.259) - (coefficient of kinetic friction x 25 x 9.8 x 0.966)
Net force = 16 + 63.26 - (coefficient of kinetic friction x 238.35)
Now, we assume that the box slides down the incline with a constant velocity (at the bottom) and hence the net force is zero.
0 = 16 + 63.26 - (coefficient of kinetic friction x 238.35)
Rearranging and solving for the coefficient of kinetic friction:
Coefficient of kinetic friction = (16 + 63.26) / 238.35 = 0.363
Therefore, the coefficient of kinetic friction between the box and the surface of the incline is 0.363.
(b) To find the work done on the box by the force of friction, we can use the formula:
Work = force x distance x cos(θ)
The force of friction can be calculated using the equation:
Friction force = coefficient of kinetic friction x normal force
The normal force is the component of the weight of the box perpendicular to the incline:
Normal force = weight x cos(θ)
Weight = mass x g
Substituting in the values:
Weight = 25 kg x 9.8 m/s^2 = 245 N
Normal force = 245 N x cos(15°)
Friction force = 0.363 x 245 N x cos(15°)
Now, we can calculate the work done:
Work = friction force x distance x cos(θ)
Note that in this case, the distance refers to the length of the incline (9 m) and the angle θ is the angle between the incline and the horizontal (15°).
Work = (0.363 x 245 N x cos(15°)) x 9 m x cos(15°)
Finally, calculate the work:
Work = 153.67 J
Therefore, the work done on the box by the force of friction is 153.67 J.
(c) The change in potential energy can be calculated using the equation:
ΔPE = mgh
Where:
m = mass of the box (25 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height of the incline
The height (h) can be calculated using the formula:
h = distance x sin(θ)
Plugging in the values:
h = 9 m x sin(15°)
Now, we can calculate the change in potential energy:
ΔPE = 25 kg x 9.8 m/s^2 x (9 m x sin(15°))
Finally, calculate the change in potential energy:
ΔPE = 315.47 J
Therefore, the change in potential energy of the box is 315.47 J.