Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]W = 121271.99 \ J[/tex]
b
[tex]v = 26.74 \ m/s[/tex]
Explanation:
From the question we are said that
The mass of the shed is [tex]m_s = 375 \ kg[/tex]
The pulling force is [tex]F = 2511 \ N[/tex]
The angle is [tex]\theta = 15^o[/tex]
The distance covered is [tex]d = 50 \ m[/tex]
The kinetic force is [tex]F_k = 255 \ N[/tex]
Generally the work done is mathematically represented as
[tex]W = F * cos(\theta ) * d[/tex]
=> [tex]W = 2511 * cos( 15 ) * 50[/tex]
=> [tex]W = 121271.99 \ J[/tex]
Generally according to the work-energy theorem
[tex]T_w = \Delta KE[/tex]
Here [tex]T_w[/tex] is total work done which is mathematically represented as
[tex]T_w = F_k * d + F d cos(\theta)[/tex]
=> [tex]T_w = 255 * 50 + 121271.99[/tex]
=> [tex]T_w = 134021.99 \ J[/tex]
Also [tex]\Delta KE[/tex] is the change in kinetic energy which is mathematically represented as
[tex]\Delta KE = \frac{1}{2} * m * [ v^2 - u^0 ][/tex]
Here u is the initial velocity of the shed when at rest
[tex]\Delta KE = \frac{1}{2} * 375 * [ v^2 - 0^0 ][/tex]
=> [tex]\Delta KE = 187.5 v^2[/tex]
So
[tex]134021.99 = 187.5 v^2[/tex]
=> [tex]v = \sqrt{714.783 }[/tex]
=> [tex]v = 26.74 \ m/s[/tex]
