Easy N2L question grade 11 physics

[CONBOSS]

Easy N2L question... grade 11 physics

Problem: A 500g dynamics cart is attached to a 50g mass, which is suspended over a pulley. When released from rest, it took 1.5 s for the cart to move a distance of 1m.

Find: a) acceleration of cart,
b) friction between cart and track




Attempt; a) Given: Δd=1m, Δt=1.5s

Δv= Δd/Δt
Δv=1/1.5
Δv=.666 m/s

a=Δv/Δt
a=.666/1.5
a=.444 m/s²

Thus, ave. acceleration of cart = .444 m/s²

b) Given: Fapp = 0.49 N, mcart=.500 kg, a=.444 m/s²

FNet= ma → (.5)(.444)
FNet= .222

FNet= Fapp + f + FN + Fg (FN & Fg are redundant)

.222= 0.49 + f → .222 - .49 = f
f= -.268 N

Therefore friction = .268 N

 

[PeterO]

Problem: A 500g dynamics cart is attached to a 50g mass, which is suspended over a pulley. When released from rest, it took 1.5 s for the cart to move a distance of 1m.

Find: a) acceleration of cart,
b) friction between cart and track




Attempt; a) Given: Δd=1m, Δt=1.5s

Δv= Δd/Δt
Δv=1/1.5
Δv=.666 m/s

a=Δv/Δt
a=.666/1.5
a=.444 m/s²

Thus, ave. acceleration of cart = .444 m/s²

b) Given: Fapp = 0.49 N, mcart=.500 kg, a=.444 m/s²

FNet= ma → (.5)(.444)
FNet= .222

FNet= Fapp + f + FN + Fg (FN & Fg are redundant)

.222= 0.49 + f → .222 - .49 = f
f= -.268 N

Therefore friction = .268 N

The line in red is incorrect. average velocity = Δd/Δt

When the object starts at rest, and has constant acceleration, the final velocity is twice the average velocity.

 

[SammyS]

Problem: A 500g dynamics cart is attached to a 50g mass, which is suspended over a pulley. When released from rest, it took 1.5 s for the cart to move a distance of 1m.

Find: a) acceleration of cart,
b) friction between cart and track

Attempt; a) Given: Δd=1m, Δt=1.5s

Δv= Δd/Δt
Δv=1/1.5 This is the average velocity. the final velocity is twice this.
Δv=.666 m/s

a=Δv/Δt
a=.666/1.5
a=.444 m/s² This is in error also.

Thus, ave. acceleration of cart = .444 m/s²

b) Given: Fapp = 0.49 N, mcart=.500 kg, a=.444 m/s²

FNet= ma → (.5)(.444)
FNet= .222

FNet= Fapp + f + FN + Fg (FN & Fg are redundant)

.222= 0.49 + f → .222 - .49 = f
f= -.268 N

Therefore friction = .268 N

See comments above.

The total amount of mass being accelerated is 550 g.

 

[CONBOSS]

Attempt; a) Given: Δd=1m, Δt=1.5s

Δv= Δd/Δt
Δv=1/1.5
Δv=.666 m/s

a=Δv/Δt
a=(.666/1.5) x 2
a=..888 m/s²

Thus, acceleration of cart = .888 m/s²

b) Given: Fapp = 0.49 N, mcart=.550 kg, a=.888 m/s²

FNet= ma → (.550)(.888)
FNet= .4882

FNet= Fapp + f + FN + Fg (FN & Fg are redundant)

.4882= 0.49 + f → .4882 - .49 = f
f= -.0016?

Therefore friction = .0016 N

 

[PeterO]

Attempt; a) Given: Δd=1m, Δt=1.5s

Δv= Δd/Δt
Δv=1/1.5
Δv=.666 m/s This is still wrong, and putting a two in the next bit does give a correct numerical answer, but is poor compensation

a=Δv/Δt
a=(.666/1.5) x 2
a=..888 m/s²

Thus, acceleration of cart = .888 m/s²

b) Given: Fapp = 0.49 N, mcart=.550 kg, a=.888 m/s²

The mass of the Cart is only 500g, not 550. The mass of the system is 550, but you are claiming only the cart here. Perhaps you mean the total mass, but you have said m_cart. You must say what you mean - which is the problem with the first part (Δv = Δd/Δt) You must state correctly, then substitute correctly



FNet= ma → (.550)(.888)
FNet= .4882

FNet= Fapp + f + FN + Fg (FN & Fg are redundant)


.4882= 0.49 + f → .4882 - .49 = f
f= -.0016?

Therefore friction = .0016 N

There is a chance this final answer is correct - but the explanations are very confused and contradictory.

 

[CONBOSS]

Above poster SammyS said total mass was 550g, so I would assume to use 550g as my total mass for finding FNet right? I will fix the question when I write it out to make it more comprehensible!

 

[PeterO]

Above poster SammyS said total mass was 550g, so I would assume to use 550g as my total mass for finding FNet right? I will fix the question when I write it out to make it more comprehensible!

You may well be correct to use 550g, but you still call it m_cart ! That's the problem.