A crate of mass 10.0 kg is pulled up arough incline with an initial speed of 1.4

Question

A crate of mass 10.0 kg is pulled up arough incline with an initial speed of 1.47 m/s. The pulling force is 93.0 N parallel to the incline, which makes anangle of 20.2° with the horizontal. The coefficient ofkinetic friction is 0.400, and the crate is pulled 4.95 m. (a) How much work is done by the gravitationalforce on the crate?
J

(b) Determine the increase in internal energy of the crate-inclinesystem due friction.
J

(c) How much work is done by the 93.0N force on the crate?
J

(d) What is the change in kinetic energy of the crate?
J

(e) What is the speed of the crate after being pulled 4.95 m?
m/s (a) How much work is done by the gravitationalforce on the crate?
J

(b) Determine the increase in internal energy of the crate-inclinesystem due friction.
J

(c) How much work is done by the 93.0N force on the crate?
J

(d) What is the change in kinetic energy of the crate?
J

(e) What is the speed of the crate after being pulled 4.95 m?
m/s

Explanation / Answer

  Initial speed  ( u ) = 1.47m/s    Force ( F ) = 93 N   Angle ( ) =20.2o    Coefficient of kinetic friction ( k ) = 0.4     Distance ( d ) = 4.95 m g = 9.8 m/s2 a )    Gravitational force on crate   F = mg                                                   = 10 kg x 9.8 m/s2                                                     = 98 N        work  W =F . S                     =   98N x 4.95m                     = 485.1 J c)     work is done by the F = 93.0 N force on the crate                         W =F . S = ----- J            d ) change inkinetic energy                 K E = K2 - K1                                    = (1/ 2) m vf2 - (1 /2 )mvi2                          = 1/2 m ( vf2 - vi2)                         = ------J               e)   Final speed of the crate    From the equation of motion         v2 = u2 + 2 as       [ a = g ]                 Solvefor v = -----m/s            Solve it I hope it helpsyou                                                                                          d ) change inkinetic energy                 K E = K2 - K1                                    = (1/ 2) m vf2 - (1 /2 )mvi2                          = 1/2 m ( vf2 - vi2)                         = ------J             d ) change inkinetic energy                 K E = K2 - K1                                    = (1/ 2) m vf2 - (1 /2 )mvi2                          = 1/2 m ( vf2 - vi2)                         = ------J               e)   Final speed of the crate    From the equation of motion         v2 = u2 + 2 as       [ a = g ]                 Solvefor v = -----m/s            Solve it I hope it helpsyou                                                                              

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.