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

   Given
       mass of crate   m =10 kg         initial speed ofcrate    u = 1.47 m/s          Force F = 93 N       angle of incline   = 20.2 o      coefficient of kinetic friction k   = 0.4       distance    d =4.95 m a ) gravitational force on crate   F =mg = 10 x 9.8 = 98 N       work is done by the gravitationalforce                    W =F . S                          =   98 x 4.95                          = 485.1 J c) work is done by the 93.0N force on the crate                         W =F . S                          =   93 x 4.95                          =  457.56 J e)   Final speed of the crate  v2 = u2 + 2 g s                                                  = 1.472 + 2 x 9.8 x 4.95                                                 v  = 9.95 m/s d ) change inkinetic energy                 K E = (1/ 2) m vf2 - (1 /2)mvi2                       = 0.5 x 10 ( 9.95 2 - 1.472 )                       =   484.208 J                          =   93 x 4.95                          =  457.56 J e)   Final speed of the crate  v2 = u2 + 2 g s                                                  = 1.472 + 2 x 9.8 x 4.95                                                 v  = 9.95 m/s d ) change inkinetic energy                 K E = (1/ 2) m vf2 - (1 /2)mvi2                       = 0.5 x 10 ( 9.95 2 - 1.472 )                       =   484.208 J

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