Question
(Total for Question is 7 marks) 3 A particle of mass 6 kg is initially at rest and is then acted upon by a force R=(ai+10j)N on a bearing of 300^circ (a) Find the exact value of a. (3) (b) Calculate the magnitude of R. (2) (c) Work out the magnitude of th acceleration of the particle. (2) (d) Find the time it takes for the particle to travel a distance of 640 m. (2)
Answer
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Grant
Elite · Tutor for 8 years
Answer
(a) Finding the exact value of "a"The given force vector has both an i and j component. The bearing of the force is 300°. We need to view from the physics perspective to answer this. In a vector, the i-component represents horizontal (east-west) motion while the j (north-south) component represents vertical motion. But in dealing with bearing, we know that bearing is measured clockwise from north, which means that a bearing of 300° acts in the Northwest quadrant.This means that our i and j components would both be negative when stated in unit vectors due to i (east-west) facing the opposite of east and j (north-south) facing opposite of north direction. Also we know in right triangles, the cosine of the angle yields the x-component and sine gives the y (when we consider the north direction as negative y due to the opposite convention of bearings).We can apply these trigonophic relation to our situation with force:Cos 300° = a/R ⇒ a = Rcos 300°Sin 300° = 10 / R Given Sin 300° = -1/2 and Cos 300°=-sqrt(3)/2, after susbtituting these values into the above equations, we can find:10 / -1/2 = r => R = -20 Rcos 300 = a -20*cos(300°)=-20*(-sqrt(3)/2)= 10 sqrt(3)Therefore, a = 10 sqrt(3).b) Calculating the magnitude of R ||R|| = √[(value of a)^2 + (value of y)^2] -> R = √[(-20)^2 + 10^2] -> R = √[400 + 100] = √500 = 10√2 N.c) Getting the magnitude of the acceleration of the particle The formula we use to get the acceleration from force is F = m*a, which gives us a = F/m-> a = ||R||/m -> a = 10√2 / 6 = approx. 2.38 m/s^2d) The time for total distance to be traveled. Firstly we need to calculate the speed at any one point in time. Speed = acceleration * time => V = at = >t = V / a.Now assuming uniform motion and a final velocity of zero (because for the case of uniformly accelerated motion from rest and return to rest while covering a certain distance, the average velocity is simply half the maximum speed due to the symmetrical nature of motion - "what goes up must come down")distance = average_velocity * time as formulas go => 640m = 1/2*V*t plugging in t = V/a from formula above => t = √(2ds/a) -> √(2X 640m / 2.38m/s^2) => = approx. 24.29 sSo we find【Answer】:(a) a = 10 sqrt(3).(b) The magnitude of the Force R = 10√58 N.(c) The magnitude of the acceleration = approx. 2.38 m/s^2.(d) The time to travel a distance of 640 m = approx. 24.29 s.