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taurus [48]
4 years ago
15

If an automobile engine delivers a power of 50.0 hp, how much time will it take for the engine to do 6.40 x 10^4 j of work? (1 h

p= 746 W)
Physics
1 answer:
aleksandrvk [35]4 years ago
7 0

Given data

Power (P) = 50 hp,

                = 50 × 746,  we know that 1 hp = 746 W.

                = 37300 Watts      (Watt = J/s)  

Work  = 6.40 ×10⁴ J

Power is defined as rate of doing work and the unit of power is<em> Watt.</em>

Mathematically,

         Power = (Work / time)   Watts

                     = 6.40 ×10⁴ / time

          37300 W = 6.40 ×10⁴ J /time      (Where time in seconds)

         => time = Work/Power

                      = 6.40 ×10⁴/37300

                      = <em>1.74 seconds  </em>

<em>  </em><em>Therefore , the engine need 1.74 seconds to do 6.40 6.40 ×10⁴ J of work </em>

<em> </em>


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The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

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(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

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(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

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<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

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<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

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