Question 29 has been answered in place of question 32. Question 32 reads: A 1.40

Question

Question 29 has been answered in place of question 32.
Question 32 reads:
A 1.40-kg block slides with a speed of 0.950 m/s on africtionless horizontal surface until it encounters a spring with aforce constant of 734 N/m. The block comes to rest aftercompressing the spring 4.15 cm. Find the spring potential energy,U, the kinetic energy of the block, K and the total mechanicalenergy of the system, E, for compressions of (a) 0 cm, (b) 1.00 cm,(c) 2.00 cm, (d) 3.00 cm, and (e) 4.00 cm.
Question 32 reads:
A 1.40-kg block slides with a speed of 0.950 m/s on africtionless horizontal surface until it encounters a spring with aforce constant of 734 N/m. The block comes to rest aftercompressing the spring 4.15 cm. Find the spring potential energy,U, the kinetic energy of the block, K and the total mechanicalenergy of the system, E, for compressions of (a) 0 cm, (b) 1.00 cm,(c) 2.00 cm, (d) 3.00 cm, and (e) 4.00 cm.

Explanation / Answer

Mass m = 1.40 kg

Speed v = 0.95 m / s

Force constant k = 734 N / m

Compression of the spring X = 4.15 cm =0.0415 m

Total energy E = ( ½) k X ^ 2

                         = 0.632 J

(a) Kinetic energy K = 0 J

Total mechanical energy E = 0.632 J

(b). Spring potential energy P = ( 1/ 2)kx^ 2

Where x = 1 cm = 0.01 m

So, P = 0.0367 J

Kinetic energy K = E – P

                             = 0.5953 J

Total mechanical energy E = 0.632 J

(c). x ‘ = 2 cm =0.02 m

Spring potential energy P = ( 1/ 2) kx ‘ ^ 2

So, P = 0.1468 J

Kinetic energy K = E – P

                             = 0.4852 J

Total mechanical energy E = 0.632 J

(d). x “ = 3 cm =0.03 m

Spring potential energy P = ( 1/ 2) kx “ ^ 2

So, P = 0.3303 J

Kinetic energy K = E – P

                             = 0.3017 J

Total mechanical energy E = 0.632 J

(e). x “' = 4 cm =0.04 m

Spring potential energy P = ( 1/ 2) kx “' ^ 2

So, P = 0.6112 J

Kinetic energy K = E – P

                             =0.0208 J

Total mechanical energy E = 0.632 J

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