Figure 9-44 shows an arrangement with an air track, in which a cart is connected

Question

Figure 9-44 shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block.The cart has mass m1 = 0.720 kg, and its center is initially at xy coordinates (?0.450 m, 0 m); the block has mass m2 = 0.300 kg, and its center is initially at xy coordinates (0, ?0.190 m). The mass of the cord and pulley are negligible. The cart is released from rest, and both cart and block move until the cart hits the pulley. The friction between the cart and the air track and between the pulley and its axle is negligible. (a) In unit-vector notation, what is the acceleration of the center of mass of the cart?block system? (b) What is the velocity of the com as a function of time t, in unit-vector notation?

Explanation / Answer

The force acting on the mass m1 in the x-direction are SFx= T + (m1a) = 0 or T = -m1a --------(1) The force acting on the mass m1 in the y-direction are SFy= n + (-w) = 0 or n + (-m1g) = 0 or n = m1g The force acting on the mass m2 in the x-direction are SFx= T + (-w) = 0 or T = m2g --------(2) From equations (1) and (2),we get m2g = - m1a or a = -(m2/m1) * g Here,m1 = 0.720 kg, m2= 0.30 kg and g = 9.8 m/s2 or a = -(0.30/0.72) * 9.8 = - 4.083 m/s2 The negative value of acceleration indicates that massm1 is being pulled by the mass m2 in the downward direction. Therefore,the acceleration of the center of mass of the cart-block system is 4.083 m/s2. The distance between the mass m1 and massm2 is S = [(0 + 0.450)^2 + (-0.190 -0)^2]^1/2 = 0.4884 m Therefore,the velocity of the common as a function of time t, in unit-vector notation is v2 - u2 = 2aS Here,u = 0 m/s or v2 = 2aS or v = (2aS)1/2 or v = (2 * 4.088 * 0.4884)1/2 = - 1.998j m/s The negative value of velocity indicates that the massm2 moves in the downward direction.

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