Question
Download Solution PDFA block of weight 2 kN is initially at rest on a rough horizontal surface (μ = 0.2). It is acted upon by a force (= P) which varies with time as shown in figure. What will be the velocity of the block at the end of 3 seconds? (assume, g = 10 m/s2)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
A block subjected to a time-varying force experiences acceleration when the applied force exceeds friction. The final velocity is calculated using Newton’s second law and kinematic equations.
Given:
Weight of block, W = 2 kN = 2000 N
Coefficient of friction, μ = 0.2
g = 10 m/s2, Time = 3 s
Force applied, P = 600 N (from graph)
Calculation:
Normal reaction, N = W = 2000 N
Frictional force, \( f = \mu N = 0.2 \times 2000 = 400~N \)
Net force, \( F_{\text{net}} = 600 - 400 = 200~N \)
Mass of the block, \( m = \frac{W}{g} = \frac{2000}{10} = 200~kg \)
Acceleration, \( a = \frac{F_{\text{net}}}{m} = \frac{200}{200} = 1~m/s^2 \)
Using the kinematic equation, \( v = u + at = 0 + 1 \times 3 = 3~m/s \)
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