Answer: v = 3.26 m/s
Explanation:
Force applied (F) = 900 N,
mass of professor and chair (m) = 90kg,
g = acceleration due to gravity = 9.8 m/s²,
Fr = frictional force = 0 ( since from the question, the rollers the chair and professor are moving is frictionless),
θ = angle of inclination = 33°
a = acceleration of object =?
From newton's second law of motion, we have that
F - (mg sinθ +Fr) = ma
Where mg sinθ is the horizontal component of the weight of the mass of professor and chair due to the inclination of the ramp.
But Fr = 0
Hence, we have that
F - mg sinθ = ma
600 - (90×9.8×sin30) = 90 (a)
600 - 480.371= 90a
119.629 = 90a
a = 119.629/ 90
a = 1.33 m/s².
But the body started the morning (at the bottom of the ramp) with a velocity of 2.30 m/s²
Hence u = initial velocity = 2.30 m/s², a = acceleration = 1.33m/s², v = final velocity =?, s = distance covered = 2m
By using equation of motion for a constant acceleration, we have that
v² = u² + 2as
v ² = (2.3)² +2(1.33)×(2)
v² = 5.29 + 5.32
v² = 10.61
v = √10.61
v = 3.26 m/s
