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PHY 101 PHY 101 Past Questions Objective Question
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Question

A load of 100 kg is placed on the elevator of mass 700 kg. Find the acceleration of the elevator when originally moving downward at 10 m/s is brought to rest with constant acceleration in a distance 20 m.
Options
A 2.5 m/s²
Correct Answer
B 2.0 m/s²
C 20 m/s²
D 25 m/s²
Correct Answer

Option A is the correct answer.

Detailed Explanation

Using v² = u² + 2as, with u = -10 m/s (downward), v = 0, s = 20 m. 0 = 100 + 2a(20) → -100 = 40a → a = -2.5 m/s². Magnitude = 2.5 m/s² upward.

Hint

v² = u² + 2as, watch sign convention.

Question Info

Course: PHY 101
Topic: PHY 101 Past Questions
Type: Objective
Difficulty: Not specified
Question 931 of 1250
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