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Sir Isaac Newton was an English mathematician and physicist - Leaving Cert Physics - Question 6 - 2020

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Sir Isaac Newton was an English mathematician and physicist. He is widely recognised as one of the most influential scientists of all time. Newton's first law of mo... show full transcript

Worked Solution & Example Answer:Sir Isaac Newton was an English mathematician and physicist - Leaving Cert Physics - Question 6 - 2020

Step 1

(i) Calculate the resultant (net) force on the 9 kg object in the diagram above. In what direction does it act?

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Answer

To find the resultant force on the 9 kg object, we need to consider all acting forces. The forces acting are 6 N to the right and 3 N to the left. The net force can be calculated as follows:

Fnet=FrightFleft=6extN3extN=3extNF_{net} = F_{right} - F_{left} = 6 ext{ N} - 3 ext{ N} = 3 ext{ N}

The direction of the resultant force is to the right.

Step 2

(ii) Calculate the acceleration of the 9 kg object.

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Answer

Using Newton's second law of motion, we can find the acceleration by dividing the net force by the mass of the object:

a=Fnetm=3extN9extkg=0.33extms2a = \frac{F_{net}}{m} = \frac{3 ext{ N}}{9 ext{ kg}} = 0.33 ext{ m s}^{-2}

Step 3

(iii) State Newton's third law of motion.

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Answer

Newton's third law of motion states that for every action, there is an equal and opposite reaction.

Step 4

(iv) Use Newton's third law to explain how a rocket takes off. (A labelled diagram may help your answer.)

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Answer

When a rocket takes off, it expels gas downwards at high speed. According to Newton's third law, this action (the downward force of gas) creates an equal and opposite reaction, pushing the rocket upwards. A labelled diagram can illustrate this by showing the rocket and the direction of the gas being expelled.

Step 5

(v) Calculate the kinetic energy of the car when it is travelling at 18 m s<sup>-1</sup>.

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Answer

The kinetic energy (KE) of an object can be calculated using the formula:

KE=12mv2KE = \frac{1}{2}mv^{2}

Substituting the known values:

KE=12×700extkg×(18extms1)2=113400extJKE = \frac{1}{2} \times 700 ext{ kg} \times (18 ext{ m s}^{-1})^{2} = 113400 ext{ J}

Step 6

(vi) Calculate the acceleration of the car.

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Answer

The acceleration (a) can be calculated using the formula:

a=Δvt=18extms106exts=3extms2a = \frac{\Delta v}{t} = \frac{18 ext{ m s}^{-1} - 0 }{6 ext{ s}} = 3 ext{ m s}^{-2}

Step 7

(vii) Calculate the net force on the car as it accelerates.

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Answer

Using Newton's second law:

Fnet=ma=700extkg×3extms2=2100extNF_{net} = ma = 700 ext{ kg} \times 3 ext{ m s}^{-2} = 2100 ext{ N}

Step 8

(viii) The engine of the car provides a driving force of 3000 N.

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Answer

The net force acts in conjunction with the driving force provided by the engine.

Step 9

(ix) Calculate the friction acting on the car.

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Answer

To find the friction force, we can use:

Ffriction=FdrivingFnet=3000extN2100extN=900extNF_{friction} = F_{driving} - F_{net} = 3000 ext{ N} - 2100 ext{ N} = 900 ext{ N}

Step 10

(x) State one method of reducing friction.

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Answer

One method of reducing friction is to use lubricants such as oil or grease on the contact surfaces.

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