A wet bar of soap slides down a ramp 5.3-m long inclined at 4.9o. How long (in seconds) does it take to reach the bottom, assume the coefficient of kinetic friction is 0.04.
To find the time it takes for the soap to slide down the ramp, we can use the principles of kinematics. The key equation we'll be using is the equation of motion for an object sliding down an inclined plane:
s = ut + (1/2)at^2
where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.
First, we need to find the initial velocity of the soap as it starts sliding down the ramp. The soap is at rest initially, so the initial velocity u is 0 m/s.
Next, we need to find the acceleration. The acceleration of an object sliding down an inclined plane can be calculated using the formula:
a = gsin(theta) - mu*gcos(theta)
where g is the acceleration due to gravity (9.8 m/s^2), theta is the angle of inclination (4.9 degrees), and mu is the coefficient of kinetic friction (0.04).
Now we can calculate the acceleration:
a = (9.8 * sin(4.9)) - (0.04 * 9.8 * cos(4.9))
Once we have the acceleration, we can now substitute the values of u (0) and a into the equation of motion:
s = ut + (1/2)at^2
5.3 = 0*t + (1/2)*a*t^2
Now we can solve this equation for t. Rearranging, we get:
(1/2)*a*t^2 = 5.3
t^2 = (2 * 5.3) / a
t = sqrt((2 * 5.3) / a)
Plug in the value of a that we calculated earlier, and solve for t to find the time it takes for the soap to reach the bottom of the ramp.
friction force=m*g*costheta*mu
gravity force= m*g*sintheta
net force=gravity-friction
= mg(sintheta-mu*costheta)
but that equals m*acceleration
so acceleration a= g(sintheta-mu*costheta)
vf^2=vi^2+2ad vi=zero
solve for vf, you have a and d
average velocity= vf/2
time = distance/averagevelocity