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2. A 5.5-kg block is initially at rest on a frictionless horizontal surface. It is pulled with a constant horizontal force of 3.8 N. (a) What is its acceleration? (1) How long must it be pulled before its speed is 5.2 m/s? (c) How far does it move in this time?? |
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Answer» Answer: Although a relatively simple PROBLEM, this is another one of those dreaded “2-part” problems that baffle many students. We will choose one of Newton’s EQUATIONS of motion shown below: S=Vit+12at2S=Vit+12at2 ————--equation 1 Vf=Vi+atVf=Vi+at —————equation 2 combine equation 1 and equation 2 to eliminate “t” gives V2f−V2i=2aSVf2−Vi2=2aS —————equation 3 The problem asks for speed after a given distance, so you would choose EITHER equation 1 or equation 3 since they both have distance ( SS ) in the equation. But since we don’t KNOW the travel time, let’s choose equation 3: V2f−V2i=2aSVf2−Vi2=2aS Before we can solve this equation, we first need to calculate the acceleration. This is why I call this a 2-part problem. F=maF=ma or a=Fm=1200N5kg=1200kg⋅ms25kg=240ms2a=Fm=1200N5kg=1200kg⋅ms25kg=240ms2 equation 3: V2f−V2i=2aSVf2−Vi2=2aS V2f−0=2(240)(5)Vf2−0=2(240)(5) Vf=49.0msVf=49.0ms |
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