A block of mass m = 2.00 kg is released from rest at h = 0.800 m from the surfac

Question

A block of mass m = 2.00 kg is released from rest at h = 0.800 m from the surface of a table, at the top of a = 35.0° incline as shown below. The frictionless incline is fixed on a table of height H = 2.00 m.

(a) Determine the acceleration of the block as it slides down the incline.
_____m/s2

(b) What is the velocity of the block as it leaves the incline?
_____m/s

(c) How far from the table will the block hit the floor?
_____m

(d) How much time has elapsed between when the block is released and when it hits the floor?
____s
(e) Does the mass of the block affect any of the above calculations?

Yes or No   

Explanation / Answer

a) F = ma, so a = F/m = 9.8*cos(35)/2kg = 4.01m/s2

b)Let length of incline = L
sin(theta) = h/L
Or L = h/sin(theta)
Or L = 0.8/sin35
Or L = 0.57 m
Initial velocity u = 0
Displacement = L = 0.57 m
Acceleration a = 4.01 m/s2 (calculated in a)
Final velocity v = ?
v2 = u2 + 2aL
Or v2 = 0 + 2*4.01*0.57
Or v2 = 4.571
Or v = 2.13 m/s
Ans: 2.13 m/s

c) Consider the motion from the bottom of the incline to the floor.
Take downward direction as positive.
In vertical direction: -
Initial velocity = v sin(theta)
Acceleration = g
Height = H
Let time taken = t
H = v sin(theta) * t + 1/2 gt2
Or 2 = 2.13 * sin(35) * t + 1/2 * 9.8 * t2
Or 2 = 0.819 t + 4.9 t2
Or 4.9 t^2 + 0.819t - 2 = 0
Or t = [-1.07 +- sqrt(1.07^2 + 4*4.9*2)]/(2*4.9)
t cannot be negative.  
Or t = 0.56 s--------------------------------(c.1)

Horizontal component of velocity = v cos(theta)
This is constant because there is no force in horizontal direction.
R = v cos(theta) * t
Or R = 2.13 * cos(35) * 0.56
Or R = 0.977m

Ans: 0.977m

d) Let t1 = time taken by the block to move to the bottom of the incline
Distance covered = L = 0.57 m (from equation (b.1))
Initial velocity u = 0
Final velocity v = 2.13 m/s (result of part b)
Average velocity vavg = (v+u)/2 = (2.13+0)/2 = 1.065 m/s
t1 = L/vavg
Or t1 = 0.57/1.565
Or t1 = 0.364 s
From equation (c.1), time taken to move from bottom of the incline to the floor = 0.56 s
Therefore, total tie taken from the time the block is released to the time it hits the floor = 0.364 + 0.56 = 0.924 s
Ans: 0.924 s

e) The mass of the block has not been used in making any of the above calculations. Therefore, it does not affect.
Ans: No

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.