The block in Figure 7-11 a lies on a horizontal frictionless surface and is atta

Question

The block in Figure 7-11a lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 65 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 3.3 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops.
Assume that the stopping point is reached.

During the block's displacement, find the following values.

Explanation / Answer

(a)Position of the block be a. now it can be stretched to the point till the spring force balances the external force. So 65 x a=3.3. so,a=3.3/65=0.0507 m (b)Work done by applied force on the block = 0.0507 x 3.3 = 0.1675 joules (c)Work done by the spring force on the block= -0.5 K a^2 [where k is the spring constant] = 0.0835 joules (d)Force on the block = 3.3 - Ka {where a is the displacement from the mean position} So acceleration is (3.3-Ka)/M. Taking it integration to get the velocity and then taking its derivative to find max we get a= 65/3.3= 19.69

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