Two blocks of mass 1 kg and 4 kg respectively are attached by a tight massless rope between them. The 1 kg block sits on the left and the 4 kg block sits on the right. The 1 kg mass has a 100 Newton force applied to it pulling it to the left. What is the acceleration of both blocks and the tension in the rope as they move across a frictionless surface?
A The acceleration is 15 m s–2 to the left and the tension is 20 N
B The acceleration is 20 m s–2 to the left and the tension is 40 N
C The acceleration is 15 m s–2 to the left and the tension is 60 N
D The acceleration is 20 m s–2 to the left and the tension is 80 N
Step 1: Consider the whole of the system
Step 2: Find the acceleration
F = m × a
a = F ÷ m
a = 100 ÷ 5 = 20 m s–2 to the left
Step 3: Examine the 4 kg mass only
F = m × a
F = 4 × 20 = 80 N to the left
Step 4: State your answer
A stationary object is subject to a 300 N force towards the left and at 55 degrees leftwards with respect to the vertical and a 450 N force to the right and 35° downwards with respect to the horizontal.Calculate what is the magnitude and direction of the third force that would make this object remain stationary (to the nearest N).
Step 1: Recall Newton's first law
A body will remain at rest or move with constant velocity unless acted on by a resultant force
Step 2: Resolve the 300 N force into its horizontal and vertical components
sin(55°) × 300 = 246 N
cos(55°) × 300 = 172 N
Step 3: Resolve the 450 N force into its horizontal and vertical components
cos(35°) × 450 = 369 N
sin(35°) × 450 = 258 N
Step 4: Combine the horizontal components
369 – 246 = 123 N to the right
Step 5: Combine the vertical components
258 – 172 = 86 N downwards
Step 6: Make a right angle triangle using these two force vectors
Step 7: Use the two vectors magnitudes to find the angle from the horizon
tan(θ°) = Opp. ÷ Adj. = 86 ÷ 123
θ° = tan–1(86 ÷ 123) = tan–1(0.699)
θ° ≈ 35°
Step 8: Use Pythagoras theorem or trigonometry to find the magnitude of the resultant force
c^{2} = 86^{2} + 123^{2} = 7396 + 15129 = 22525
c = √22 525 ≈ 150 N
cos(35°) = Adj. ÷ Hyp. = 123 ÷ Hyp.
123 ÷ cos(35°) = Hyp. ≈ 150 N
Step 9: State the final answer
You are expected to know Newton's three laws of motion from memory and how they apply to physical situations. So be sure to practice and use them without having to review them before carrying out a problem.
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