2nd attempt at having this answered correctly please only answer if you are sure

Question

2nd attempt at having this answered correctly please only answer if you are sure.... In the figure below, a block of mass m = 17 kg is released from rest on a frictionless incline angled of angle = 30°. Below the block is a spring that can be compressed 2.0 cm by a force of 280 N. The block momentarily stops when it compresses the spring by 6.2 cm. Set gravitational potential energy of the block-Earth system equal to zero when the block momentarily stops on the compressed spring.

(a) As the block slides down the ramp, does the mechanical energy of the block-Earth-spring system increase, decrease, or remain the same?

Remain the same Decrease     Increase

(b) What is the elastic potential energy of the spring when it is compressed 6.2 cm?
26.91 J

(c) What is the gravitational potential energy of the block-Earth system when the block is released?
26.91 J

(d) How far does the block move down the incline from its rest position to its stopping point?
???? m the answer is not 32.3

(e) What is the kinetic energy of the block just as it touches the spring?
????? J answer is not 26.91 J

Explanation / Answer

a)

as per conservation of energy , energy is neither created nor destroyed. it only converts from one type of energy to another. since the inclince surface is frictionless, no energy is lost and hence

the mechanical energy of the block-Earth-spring system remain the same.

b)

spring force = Fs = 280 N

xs = compression of spring = 2 cm = 0.02 m

spring constant is given as

k = Fs/xs = 280/0.02 = 14000 N/m

x = compression of spring when the block stops = 6.2 cm = 0.062

spring potential energy is given as

Es = (0.5) k x2 = (0.5) (14000) (0.062)2 = 26.91 Joules

c)

using conservation of energy

Potential energy of block when released = spring potential energy when block stops = Es = 26.91 Joules

d)

d = distance moved parallel to inline

h = height dropped = d Sin30

using conservation of energy

Potential energy of block when released = spring potential energy when block stops = Es = 26.91 Joules

mgh = 26.91

mgd Sin30 = 26.91

(17 x 9.8 ) (0.5) d = 26.91

d = 0.323 m

e)

d' = distance travelled parallel to incline before touching the spring = d - compression of spring = 0.323 - 0.062 = 0.261 m

h' = height dropped = d' Sin30

using conservation of energy

kinetic energy as the block touch the spring = potential energy lost

KE = mgh' = mgd' Sin30

KE = 17 x 9.8 x 0.261 Sin30

KE = 21.74 J

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