A block of mass 5 kg is sliding down a smooth inclined plane as shown in the figure. The spring arranged near the bottom of the inclined plane has a force constant 600 N/m. Find the compression in the spring at the moment the velocity of the block is maximum?
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Answered by
6
Hey mate,
● Answer- 0.05 cm
● Explanation -
Here, Force on block due to spring
Fs = mgsinθ
Fs = 5×10×3/5
Fs = 30 N
Restoring force here is
Fr = -kx
Fr = -600×x
Here, restoring force equlizes to force on block thus,
Fs = -Fr
30 = -(-600x)
x = 30/600
x = 0.05 cm
Compression is 0.05 cm when velocity is max.
Hope that is useful...
● Answer- 0.05 cm
● Explanation -
Here, Force on block due to spring
Fs = mgsinθ
Fs = 5×10×3/5
Fs = 30 N
Restoring force here is
Fr = -kx
Fr = -600×x
Here, restoring force equlizes to force on block thus,
Fs = -Fr
30 = -(-600x)
x = 30/600
x = 0.05 cm
Compression is 0.05 cm when velocity is max.
Hope that is useful...
Answered by
5
HEYA MATE HERE U GO.
Given that
m= 5kg
k= 600N/m
g = 10m/s^2
compression x = ?
From the diagram , sin theta = 3/5
From the Newtonś third law , force due to block on spring FB = - Restoring force of spring FR
Hence , FB = FR ===> mgsintgeta = Kx
==> 5× 10 × 3/5 = 600 × x
==> 5 × 10× 3/5 = 600 × x
==> x = 30/600
==> 0.05m
==> 5 cm.
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