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An iron block of mass m = 500 kg is kept at the back of a truck moving at a speed v_(0)=90km h^(-1). The driver applies the brakes and shows down to a speed of v=54km h^(-1) in 10s. What constant force acts on the block during this time if the block does not slide on the truck-bed? |
| Answer» <html><body><p></p>Solution :The acceleration, `a=(v-v_(0))/(t)` <br/> `=(54 km h^(-1)-90kmh^(-1))/(10s)` <br/> `=(15ms^(-1)-25ms^(-1))/(10s)` <br/> `=-1m//s^(2)` <br/> or, Force F = ma <br/> `=500kg (-1 ms^(-2))` <br/> `=-500N` <br/> -ve sign <a href="https://interviewquestions.tuteehub.com/tag/indicates-1041501" style="font-weight:bold;" target="_blank" title="Click to know more about INDICATES">INDICATES</a> that the force <a href="https://interviewquestions.tuteehub.com/tag/acts-848461" style="font-weight:bold;" target="_blank" title="Click to know more about ACTS">ACTS</a> opposite to the <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> of the block. The magnitude of force is <a href="https://interviewquestions.tuteehub.com/tag/500-323288" style="font-weight:bold;" target="_blank" title="Click to know more about 500">500</a> N.</body></html> | |