a barrel of oil at rest has a mass of 65kg it is rolled down a 1.2m ramp. Due to constant force exerted on it, the barrel reaches a velocity of 4.2 m/s upon leaving the ramp. Find the constant force on the barrel while it is on the ramp?
Help please.
It is unclear whether the barrel is rolling or sliding. If one assumes sliding down a frictionless ramp,
F=ma
where vf^2=2a*4.2 or a=vf^2/8.4
solve for f.
Now, if rolling, come back and say so, it is more complicated.
rolling down a ramp not sliding
Now if it is rolling, the filled barrel has a moment of inertia of 1/2 Mr^2
The angular velocity at the bottom must be w=Vf/r
KE at bottom=PE at top
1/2 I w^2+ 1/2 m vf^2=mgh
so solve for h.
To find the constant force exerted on the barrel while it is on the ramp, we can use Newton's second law of motion, which states that the force exerted on an object is equal to its mass multiplied by its acceleration.
In this case, we need to first calculate the acceleration of the barrel as it rolls down the ramp. We can use the kinematic equation for acceleration:
v^2 = u^2 + 2as,
where:
v = final velocity = 4.2 m/s,
u = initial velocity = 0 m/s (since the barrel is at rest),
a = acceleration (to be determined),
s = distance travelled down the ramp = 1.2 m.
Rearranging the equation, we find:
a = (v^2 - u^2) / (2s),
= (4.2^2 - 0^2) / (2 * 1.2),
= 17.64 / 2.4,
= 7.35 m/s^2.
Now that we know the acceleration, we can calculate the force using Newton's second law:
F = ma,
= (mass of the barrel) * a,
= 65 kg * 7.35 m/s^2,
= 477.75 N.
Therefore, the constant force exerted on the barrel while it is on the ramp is 477.75 Newtons.