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thinking and inquiry 1. a stationary box of mass 4.2 kg is given a push…

Question

thinking and inquiry

  1. a stationary box of mass 4.2 kg is given a push of 8.2 n s along a surface where the frictional force acting is 5.8 n n. the push lasts for 3.6 s and then the box is allowed to slide on its own until it comes to rest. (10 marks)

(a) draw free-body diagrams to show the box being pushed and sliding on its own.
(b) determine the acceleration of the box as it is being pushed.
(c) calculate the speed of the box just as the push ceases.
(d) determine the acceleration of the box as it is sliding on its own.

Explanation:

Part (a)

Step1: Identify forces (pushed)

When pushed, forces: Push (\(F_{push} = 8.2\,\text{N}\) [S]), Friction (\(F_f = 5.8\,\text{N}\) [N]), Weight (\(F_g = mg\)) downward, Normal (\(F_N = F_g\)) upward.

Step2: Draw free - body (pushed)

  • Vertical: \(F_N\) (up) and \(F_g\) (down), equal in magnitude.
  • Horizontal: \(F_{push}\) (S) and \(F_f\) (N), \(F_{push}>F_f\).

Step3: Identify forces (sliding)

When sliding, push is removed. Forces: Friction (\(F_f = 5.8\,\text{N}\) [N]), Weight (\(F_g\)) downward, Normal (\(F_N = F_g\)) upward.

Step4: Draw free - body (sliding)

  • Vertical: \(F_N\) (up) and \(F_g\) (down), equal.
  • Horizontal: Only \(F_f\) (N) opposing motion.
Part (b)

Step1: Net force (pushed)

Use \(F_{net}=F_{push}-F_f\). \(F_{push} = 8.2\,\text{N}\), \(F_f = 5.8\,\text{N}\), so \(F_{net}=8.2 - 5.8=2.4\,\text{N}\) [S].

Step2: Acceleration (pushed)

Use \(F = ma\), so \(a=\frac{F_{net}}{m}\). \(m = 4.2\,\text{kg}\), \(a_{pushed}=\frac{2.4}{4.2}\approx0.571\,\text{m/s}^2\) [S].

Step3: Net force (sliding)

When sliding, net force is friction (opposing motion), \(F_{net}=-F_f=- 5.8\,\text{N}\) (N direction).

Step4: Acceleration (sliding)

\(a_{sliding}=\frac{F_{net}}{m}=\frac{- 5.8}{4.2}\approx - 1.381\,\text{m/s}^2\) (N direction, deceleration).

Part (c)

Step1: Use kinematic equation

Use \(v = u+at\). Initial velocity \(u = 0\), \(a = a_{pushed}\approx0.571\,\text{m/s}^2\), \(t = 3.6\,\text{s}\).

Step2: Calculate speed

\(v=0 + 0.571\times3.6\approx2.056\,\text{m/s}\) [S].

Part (d)

Answer:

s:
(a) Free - body diagrams:

  • Pushed: Vertical - \(F_N\) (up), \(F_g\) (down); Horizontal - \(F_{push}\) (S), \(F_f\) (N).
  • Sliding: Vertical - \(F_N\) (up), \(F_g\) (down); Horizontal - \(F_f\) (N).

(b) Acceleration (pushed): \(\approx0.57\,\text{m/s}^2\) [S]; Acceleration (sliding): \(\approx - 1.38\,\text{m/s}^2\) (N).

(c) Speed: \(\approx2.06\,\text{m/s}\) [S].

(d) Acceleration: \(\approx - 1.38\,\text{m/s}^2\) (N, or \(1.38\,\text{m/s}^2\) deceleration).