Question #217377

Use the information below to answer the following 4 questions.

A 10.0-kg object is accelerated horizontally from rest to a velocity of 11.0 m/s in 5.00 s by a horizontal force.


1.The acceleration of the object is ___m/s2. (Give your answer with 3 sig digs and do not include units).


2.The horizontal force given to the object is ___N. (Give your answer with 3 sig digs and do not include units).


3.The distance travelled by the object is ___m. (Give your answer with 3 sig digs and do not include units).


4.The work done on the object is ___J. (Give your answer with 3 sig digs and do not include units).


1
Expert's answer
2021-07-16T08:52:05-0400

1.By defintion, the magnitude of the acceleration is given as follows:


a=vfvita = \left| \dfrac{v_f - v_i}{t} \right|

where t=5st = 5s is the time taken by the object to thange its velocity from the initial one vi=0v_i = 0 to the final one vf=11m/sv_f = 11m/s. Thus, obtain:


a=11m/s5s=2.20m/s2a = \dfrac{11m/s}{5s} = 2.20m/s^2

2. According to the second Newton's law, the force acting on the object is given as follows:


F=maF = ma

where m=10kgm = 10kg is the mass of the object. Thus, have:


F=10kg2.2m/s2=22.0NF = 10kg\cdot 2.2m/s^2 = 22.0N

3. The distance travelled by the object under the constant acceleration (and zero initial speed) is given by the kinematic equation:


d=at22=2.2m/s2(5s)22=27.5md = \dfrac{at^2}{2} = \dfrac{2.2m/s^2\cdot (5s)^2}{2} = 27.5m

4. The work done on the object is given as follows:


W=Fd=22.0N27.5m=605JW = Fd = 22.0N\cdot 27.5m = 605J

Answer. 1) 2.20, 2) 22.0, 3) 27.5, 4) 605.


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