A block of mass m = 2.00 kg is released from rest h = 0.700 m from the surface o

Question

A block of mass m = 2.00 kg is released from rest h = 0.700 m from the surface of a table, at the top of a = 25.0° incline as shown in Figure P5.58. The frictionless incline is fixed on a table of height H = 2.00 m.

Please jusst solve c, d, and e and show work.

(a) Determine the acceleration of the block as it slides down the incline. 4.15 m/s2

(b) What is the velocity of the block as it leaves the incline? 3.71 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?

Explanation / Answer

  a)

the force acting to accelerate the block will be mgsin
the acceleration will be that force divided by the mass a = F/m
so the acceleration will be
a = gsin
a = (9.81)sin 25
a = 4.15 m/s/s

b)

from conservation of energy

mgh = mv^2/2
v = (2gh)^.5
v = (2(9.81)(0.7))^0.5
v = 9.81^0.5
v = 3.71 m/s

c)

R = vt
t = R/v
H = (1/2)gt^2
2H/g = (R/v)^2
from part b above v^2 = g
so the horizontal component is gcos25
2H/g = R^2/gcos25
2Hgcos25/g = R^2
R = (2Hcos25)^0.5
R = (2(2)(0.906))
R = 1.90 m

d)

ramp time from v = at
t = 3.71/4.15
t = 0.893 sec
free fall time from the vertical position equation
h = vt +gt^2/2
h = vt +gt^2/2
2 = 3.71sin25t + 9.81(t^2)/2
0 = 4.905t^2 +1.56t - 2
from quadratic equations
t = 0.500 or -0.817
as a negative time does not make sense
t = 0.500 seconds

so total time elapsed =0.500 + 0.500 = 1 seconds

(e)

No, the mass does not affect any result since it does not appear in any of the above calculations.

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