forces and elasticity (topic 5)

Question 1

describe a method the student could have followed to obtain this data of spring constant (4)

Answer 1

• measure original length of spring using a ruler
• attach spring to clamp stand and suspend a 200 g mass from the spring
• record new length using a ruler
• extension = new length - original length
• increase the mass in 200 g increments up to a total of 1 kg

Question 2

explain what is meant by non-elastic deformation (1)

Answer 2

the spring has been permanently deformed

Question 3

the mass of the student is 60 kg, calculate the weight of the student, gravitational field strength on Earth is 9.8 N/kg

Answer 3

• weight = mass x gravitational field strength
• weight = 60 x 9.8 = 588 N

Question 4

explain why the bungee company chose cord A instead of cord B (2)

Answer 4

• bungee cord B undergoes plastic deformation, would not go back to its original length
• meaning they would need to be replaced much more frequently than bungee cord A

Question 5

describe the relationship between force and extension (2)

Answer 5

• directly proportional
• or as force increases, extension increases
• in a linear way

Question 6

explain what is meant by elastic behaviour (1)

Answer 6

when the force is removed, the spring returns to its original length

Question 7

explain which student would obtain a more accurate value for the length of the spring (2)

Answer 7

• student A results will not be accurate as the ruler is not vertical
• student B would get a more accurate reading
• the set square allows the student to ensure the ruler is vertical

Question 8

calculate the force when the mass is 300g

Answer 8

• force = mass x gravitational field strength
• 300 g = 0.3 kg
• force = 0.3 x 9.8 = 2.94 N

Question 9

Answer 9

Question 10

state Hookes law (1)

Answer 10

force is directly proportional to the extension of the spring

Question 11

the car has suspension springs which have a spring constant of 20,000 N/m, the car has a mass of 1,500 kg and is supported by 4 suspension springs, calculate the compression of each spring while the car is stationary

Answer 11

Question 12

when the car travels over a speed bump, an additional force of 3,000 N is applied to each suspension spring, calculate the additional compression of the spring due to the speed bump, spring constant = 20,000

Answer 12

• extension = force ÷ spring constant
• extension = 3,000 ÷ 20,000 = 0.15 m

Question 13

describe how the comfort of the ride would vary if the spring constant of the suspensions springs was significantly increased (2)

Answer 13

• suspension springs would not compress as much with a higher spring constant
• ride would be more uncomfortable
• higher force exerted on passengers due to stiffer springs

Question 14

calculate the useful power output of the student, work done = 2.45 J, closed hand 20 times in one minute

Answer 14