Solution:
f ( t )= 20 S ( t ) + 55/30 tS ( t )− 55/30 ( t − 30 ) S ( t − 30 )
• Taking the Laplace Transform:
F ( s ) = 20/s + 55/30 ( 1/s^2 ) – 55/30 ( 1/s^2) e^-30s = 20/s + 55/30 ( 1/s^2 ) ( 1 – e^-30s)
Answer: Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside
Explanation:
Given that;
friction force of ground box = 68 N
student of 7th grade = n
Whitmore can apply a force of 25 N
every other 7th grade student can apply a force of 6 N.
now
friction force = forced applied by whitmore + total force ny 7th grade student
we substitute
68 = 25 + 6n
6n = 68 - 25
6n = 43
n = 43/6
n = 7.17
Therefore Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside