Question

• 30 In Fig. 7-10a, a block of mass m lies on a horizontal frictionless surface and is attached to one end of a horizontal spring (spring constant k) whose other "O 0.5 1 1.5 2 end is fixed. The block is initially at x (m) rest at the position where the spring Figure 7-36 Problem 30. is unstretched (x = 0) when a con- I stant horizontal force É in the positive direction of the x axis is applied to it. A plot of the resulting kinetic energy of the block versus its position x is shown in Fig. 7-36. The scale of the figure's vertical axis is set by Ks = 4.0 J. (a) What is the magnitude of É? (b) What is the value of k?

Answer 1

Graph is between displacement (1) and Kinetic enesgy of the block K(J). At x=0, Ki=OJ At K₂=1m, K₂ = ks =4.05 At m=zm, Kg = OJ potential energy stored in the spring ů P. E = ļ kx² where, kis the spring constant and a is the spring strenched length. work done on the block of mass m due to constant force f in W= (F) (x) Constant force on the block will Change its kinetic energy and strech the spring connected with it.

Applying the energy conservation work done on the block = Change in Kinetic energy of block + Change in potential energy of spring. Apply energy conservation between * = 1m and 2, com F(x2-x) = (K-kg) + (2-3) Flam-om) = (4.0J - OT) +4k (C2mo?-comp) F(m) = 4.0 (5) +įk (m2). (1) Note, Here m² is mette) and m te meter unit

Apply enesgy conservation between Xz=2m and arom Fl2z-) (K3-K.) + į klaž=x2) F( 2m-om) = (01-05) +įk (emB=(OMB) 2F (m) = 2 k (m2) K (m2) = f(m) --- (2) Note, Here m² is metre) 2 and n is meter unit.

Substitute eguation (2) in equation (1) Flm) = 4.0(5) + Ź Flm) Flm) - į f(m) = 4.0(5) F = (3)(4-0) F = 18.00 --- (3) in equation (2) Past (b) Substitute value of f=8.on km²) = f(m) k (m2) = f.ON (M) k=8.0 Nm