gram, the acceleration of the mass is 3.25 m/s.(Assume that small that it can be

Question

gram, the acceleration of the mass is 3.25 m/s.(Assume that small that it can be ignored.) When the same magnitude force is another object, the acceleration is 2.75 m/s?. What is the mass of this obiecto What would the second object's acceleration be if a force twice as large were applied to it? Show your calculations friction is 4. When a force is applied to an object with mass equal to the standard kilo. so Given an object with mass equal to the standard kilogram, how would vou determine if a force applied to it has magnitude equal to one newton? (As sume that frictional forces are so small that they can be ignored.) 5. In Question 6, assume that friction is so small that it can be ignored. 6. The spring scale in the diagram below reads 10.5 N. If the cart moves toward the right with an acceleration also toward the right of 3.25 m/s, what is the mass of the cart? Show your calculations and explain. In Questions 7-9, friction may not be ignored. 7. The force applied to the cart in Question 6 by spring scale FA is still 105N The cart now moves toward the right with a constant velocity. What are t

Explanation / Answer

(4). Mass is equal to standard kilogram , so m =1 kg

Given Acceleration a = 3.25 m/s2

so Force F = MAss * Acceleration

= 1 * 3.25

= 3.25 N

Same Force is applied to another body which causes the body to accelerate at a rate = 2.75 m/s2

Therefore mass of that body =Force / a

=3.25/2.75

Answer = 1.182 kg

if the same body was given twice the amount of force

Acceleration = Force /mass = 2*3.25/1.182

ANS = 5.5m/s2

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(5). Force = mass /Acceleration

One Newton is the force required to make a mass of standard kilogram accelerate at a rate of 1 m/s2.

so if we can mesaure the acceleration of the bodyof mass 1 kg to 1m/s2 the the force applied to it is 1N

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(6). Given Force is applied to right side

Value of force is 10.5 N

and Acceleration towards right is 3.25m/s2

Find mass.

Mass m = Force/ acceleration

=10.5/3.25

ANS= 3.23 kg

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(7). Given FA = 10.5N and velocity is constant

that means there is no acceleration , a=0 m/s2

Now frictional force always acts oppsite to direction of movement so in this case, towards LEFT

So DIRECTION is opposite to FA (towards LEFT)

Resultant force is FA - Ffriction

FA - Ffriction = mass * acc. (given a =0)

FA - Ffriction =0

FA = Ffriction = 10.5 N

magnitude and direction of frictional force is 10.5 N towards left

Note: when FA becomes larger than Ffriction the body starts accelerating

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(8).Given FA = 10.5 N(towards right) and acceleration towards right is 1.75 m/s2

Find Ffriction

Resultant force is FA - Ffriction (Ffriction is opposite to FA)

FA - Ffriction = mass * acc.

From Q(6) we found that mass of cart = 3.23 kg

10.5 -Ffriction = 3.23 * 1.75

Ffriction = 10.5 - 3.23*1.75

ANS=4.8475 N (towards left)

Note: Applied force is greater than frictional force so the body is accelerating.

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(9). Given FA = 10.5 N

Constant velocity so Acc. is 0 m/s2

Frictional force is same as in Q(8)

Resultant force is FA -FB -Ffriction

FRE = mass * acc. = 0

FA -FB -Ffriction =0

so FB = FA - Ffriction = 10.5 - 4.8475

ANS= 5.6525 N

The Spring Scale will show a value of 5.6525 N

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