Question:
A block of mass 5 kg is sliding with constant velocity of \( 6 \, ms^{-1} \) on a frictionless horizontal surface. The force exerted on the horizontal surface is:
A block of mass 5 kg is sliding with constant velocity of \( 6 \, ms^{-1} \) on a frictionless horizontal surface. The force exerted on the horizontal surface is:
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The horizontal velocity value of \( 6 \, ms^{-1} \) is extra data meant to distract you. Because the surface is frictionless and speed is completely uniform, no horizontal force component acts on the block.
Updated On: Oct 6, 2026
- 30 N
- 1.2 N
- 150 N
- 49 N
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The Correct Option is D
Solution and Explanation
Concept:
Since the block moves with a uniform velocity along the horizontal surface, the net acceleration along the horizontal plane is zero. Along the vertical axis, the block remains in static equilibrium, meaning the normal upward reaction force from the surface perfectly balances the weight of the block.
Step 1: Applying vertical force equilibrium.
The weight of the object acting downward onto the surface is: \[ W = m \cdot g \] Using mass \( m = 5 \, \text{kg} \) and standard gravitational acceleration \( g = 9.8 \, ms^{-2} \): \[ W = 5 \times 9.8 = 49 \, \text{N} \]
Step 2: Applying Newton's third law.
By Newton’s third law of action-reaction, the force exerted by the block downward on the horizontal platform matches the upward normal reaction force. Thus, the normal force exerted onto the ground is exactly \( 49 \, \text{N} \).
Step 1: Applying vertical force equilibrium.
The weight of the object acting downward onto the surface is: \[ W = m \cdot g \] Using mass \( m = 5 \, \text{kg} \) and standard gravitational acceleration \( g = 9.8 \, ms^{-2} \): \[ W = 5 \times 9.8 = 49 \, \text{N} \]
Step 2: Applying Newton's third law.
By Newton’s third law of action-reaction, the force exerted by the block downward on the horizontal platform matches the upward normal reaction force. Thus, the normal force exerted onto the ground is exactly \( 49 \, \text{N} \).
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