1) The figure(Figure 1) shows the velocity graph of a 2.5 kg object as it moves

Question

1) The figure(Figure 1) shows the velocity graph of a 2.5 kg object as it moves along the x-axis. What is the net force acting on this object at t= 1 s?

2)It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s . A 63 kg passenger gets aboard on the ground floor.

A)What is the passenger's weight before the elevator starts moving?

B)What is the passenger's weight while the elevator is speeding up?

C)What is the passenger's weight after the elevator reaches its cruising speed?

3)A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 35 angle. The block's initial speed is 10 m/s . The coefficient of kinetic friction of wood on wood is k=0.200.

A)What vertical height does the block reach above its starting point?

B)What speed does it have when it slides back down to its starting point?

Explanation / Answer

1)    Fnet = m x a (Newton's 2nd Law)

at t = 1, a = 6 m/s/s (gradient of graph at that point)

Therefore F = 2.5 x 6 = 15 Newtons (in direction of acceleration)

2)   V = a t
10 = a x 4.0
a = 2.50 m/sec^2

What is the passenger's weight before the elevator starts moving?
F = m g = 9.8 x 63 = 617.4 N

What is the passenger's weight while the elevator is speeding up?
F = m (g + a) = (9.8 + 2.50) x 63 = 79.07 N

What is the passenger's weight after the elevator reaches its cruising speed?
F = m g = 9.8 x 63 = 617.4  N

3)    The block’s acceleration up the plane from Newton’s 2nd law is:

F(x) = ma = -mgsin - mgcos
a = -g(sin + cos)
= -9.8m/s²(sin30° + 0.20cos30°)
= -6.6m/s²

(a) When the block reaches its maximum height, it s velocity is zero (v = 0), while initial velocity, v equals 10m/s. So find the distance it moves parallel to the incline, and then use trig to find the vertical height it achieves. The parallel distance is:

v² = v² + 2ax
x = -v² / 2a
= -(10m/s)² / 2(-6.6m/s²)
= 7.6m

Then from trig, the vertical height is:

sin = y / x
y = xsin
= 7.6msin35°
= 4.35m

(b) The acceleration changes for the block as it slides down the incline because now gravity works opposite of friction:

F(x) = ma = mgsin - mgcos
a = g(sin - cos)
= 9.8m/s²(sin35° - 0.20cos35°)
= 4.01m/s²

So its speed when it reaches the bottom of the incline is (remember, the parallel distance it travels is still 7.6m):

v² = v² + 2ax
= 0 + 2(4.01m/s²)(7.6m)
v = 7.80m/s

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