36 One dimensional Motion (a) its speed att 2: (b) its acceleration at 4 ; (c) t

Question

36 One dimensional Motion (a) its speed att 2: (b) its acceleration at 4 ; (c) the average acceleration between t 1s andt3s. 2.5 A particle of mass 1.5 kg has an initial velocity of (1+ Sj) m 5-1. It is acted on by a rei + 15(j-5t2k where me t2 s, and force of (4i-j) N for 3s. What is its final velocity 2.6 A particle of mass 2kg is observed to follow a trajectory r i j and k are orthogonal unit vectors. Calculate the particle speed at t the force on the particle. Describe the motion 2y at rest, is pushed along a horizontal table top by a horizontal force of 2N. After 1s it has been moved 1.5m. The force of friction is 0.4N (a) How much work is done on the book by the 2N force? (b) How much work is done on the book by frietion? (c) What is the book's kinetic energy after 1 s? (d) What power is required? 2.8 If 10kW are required to drive a 1000kg car at 72kmh 1 on the level, what is the total retarding force? what power is required to drive at T2 kmh-1 up a 10% gradient? What is the gradient of the hill if the car coasts (Gie. uses no power) down at a constant 72km h2 2.9 A bullet of mass m, traveling horizontally at speed u, strikes and comes to rest in a stationary wooden block of mass M. The block is suspended as shown (Figure 2.10). Obiain an expression for the maximum beight h, attained by the block after absorbing the impact of the bullet. Figure 2.10 Figure for Problem 2.9

Explanation / Answer

W = F.d = F d cos(theta)

d = 1.5 m

(A) F = 2 N , theta = 0 deg

W = 2 x 1.5 x cos0 = 3 J

(B) F = 0.4 N, theta = 180 deg

W = 0.4 x 1.5 x cos180 = -0.6 J  

(C) total work done = change in KE

3 + (-0.6) = K - 0

KE = 2.4 J  

(D) P = W / t = 2.4 J / 1s = 2.4 W

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