A constant horizontal force F is applied to a 5 kg box on a frictionless horizontal surface. The box starts from rest and moves
1 answer:
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1) the weight of an object at Earth's surface is given by 
, where m is the mass of the object and 
is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is
2) On Mars, the value of the gravitational acceleration is different:
. The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: 
3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:
4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:
5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as
<span>6) On Earth, the gravity acceleration is </span><span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is
</span>
<span>
</span>
2) On Mars, the value of the gravitational acceleration is different:
3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:
4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:
5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as
<span>6) On Earth, the gravity acceleration is </span>
</span>
</span>
Answer:
accelerating
Explanation:
If we consider(v > u) Acceleration:
final velocity(v)= 14m/s
initial velocity(u)=10m/s
time taken(t)= 2 seconds
a= =2m/s²
If we consider (v<u) Deceleration:
final velocity(v)= 3m/s
initial velocity(u)=9m/s
time taken(t)=2 seconds
a= = -3m/s²