A wooden box of mass 10. kg slides down a ramp from a height of 1.0 m onto a frictionless tabletop. It is brought to rest by a compressed spring. The spring constant is 1250 N/m. a) Calculate the maximum distance the spring is compressed. b) Determine the speed and acceleration of the block when the spring is compressed a distance of 14 cm.

Answers :

Part (a) Answer : The maximum compression in the spring will be : x_max = 0.3962 m = 39.62 cm

Part (b) Answer : The speed and the acceleration of the block will be :

Speed : v = 4.14 m/s

Acceleration : a = 17.5 m/s²

Step-by-step explanation

Here we have given :

Mass of the block : m = 10 kg

Height of the ramp : h = 1 m

And, Spring constant : k = 1250 N/m

Let the maximum compression in the spring be x_max.

Now, According to the conservation of energy : The potential energy stored in the spring at the maximum compression of the spring will be equal to the initial gravitational potential energy of the spring.

∴ (1/2) k (x_max)² = m g h

∴ (1/2)(1250 N/m) (x_max)² = (10 kg)(9.81 m/s²)(1 m)

∴ (x_max)² = 0.15696 m²

∴ x_max = 0.3962 m = 39.62 cm

Therefore, The maximum compression in the spring will be : x_max = 0.3962 m = 39.62 cm

Part (b) Solution :

Here, Compression in the spring : x = 14 cm = 0.14 m

Now, According to the conservation of energy : PE_spring + KE_block = GPE_block

∴ (1/2) k x² + (1/2) m v² = m g h

∴ (1/2) m v² = m g h - (1/2) k x²

∴ (1/2) m v² = (10 kg)(9.81 m/s²)(1 m) - (1/2)(1250 N/m)(0.14 m)²

∴ (1/2) m v² = 85.85 J

∴ (1/2)(10 kg) v² = 85.85 J

∴ v² = 17.17 m²/s²

∴ v = 4.14 m/s

And, When the spring is compressed : The force acting on the block will be : F_net = F_spring = k x

Since, F_net = m a

∴ m a = k x

∴ (10 kg) a = (1250 N/m)(0.14 m)

∴ a = 17.5 m/s²

This acceleration will be negative acceleration : Because the spring will be opposing the motion of the block.

Please see the attached file for the complete solution