1.2 m H 650 N/m 4.0 kg 0.80 m Target Note: Figure not drawn to scale. Block A of

Question

1.2 m H 650 N/m 4.0 kg 0.80 m Target Note: Figure not drawn to scale. Block A of mass 4.0 kg is on a horizontal, frictionless tabletop and is placed against a spring of negligible mass and spring constant 650Nm. The other end of the springis attached to a wall. The block is pushed toward the wall until the spring has been compressed a distancex, as shown above. The block is released and follows the trajectory shown, falling 0.80 m vertically and striking a target on the floor that is a horizontal distance of 1.2 from the edge of the table. Air resistance is negligible. (a) Calculate the time elapsed from the instant block A leaves the table to the instant it strikes the floor. (b) Calculate the speed of the block as it leaves the table. (c) Calculate the distance x the spring was compressed.

Explanation / Answer

Force acted on the block when it is compressed to distance x = Kx

spring constant ,k = 650 Nm

so, Force acted = 650x

So, acceleration acted, a =F/m =650x/4   m/s2

applying formula

v2 - u2 = 2as

initial velocity,u =0

acceleration =650x/4   m/s2

Velocity after block leaves the table ,v= sqrt(2*650*x2/4)

                                                            =18.027x                         ------------------------(1)

Let time taken by block to reach ground =t

Distance target hit the ground from edge of table = 1.2 m

=>   1.2 = 18.027*x*t                        (By, Distance= speed * time)      -----------(2)

Applying vertical formula :-

S=ut + 1/2*g*t2

Vertical component of velocity is zero when block leaves the table

=>   0.8 = 0.5 *9.8 *t2

=>    t = 0.404 sec

Putting value of t in equation (2)

1.2 = 7.284*x

=>   x = 0.1647 m

Putting value of x in equation (1)

Velocity after block leaves the table ,v = 18.027*0.1647 m/sec

                                                             =2.969 m/sec

(a)   Time elapsed block leaves the table to instant it strikes the floor = 0.404 sec

(b)    Speed of the block as it leaves the table = 2.969 m/sec

(c)     Distance x spring was compressed = 0.1647 m

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