Question

test 3 free response

1. a sled of mass 40.17 kg is pulled to the right along a snow - covered surface with a force of 21.98 n at an angle of 57.29°. the free - body diagram for the sled is shown below.

(a) what is the pulling force in the x direction? what is the x component of the pulling force?
(b) what is the pulling force in the y direction? what is the y component of the pulling force?
(c) what is the weight of the sled?
(d) what is the normal force from the ground on the sled?
(e) what is the acceleration of the sled?

Explanation:

Step1: Calculate the x - component of the pulling force

The formula for the x - component of a force \(F\) at an angle \(\theta\) is \(F_x = F\cos\theta\). Given \(F = 21.98\space N\) and \(\theta=57.29^{\circ}\), we have \(F_x=21.98\times\cos(57.29^{\circ})\). Using a calculator, \(\cos(57.29^{\circ})\approx0.54\), so \(F_x = 21.98\times0.54 = 11.87\space N\) (the original hand - written answer used the wrong value for the force magnitude in the formula, it should be \(F = 21.98\space N\) not \(40.17\space kg\)).

Step2: Calculate the y - component of the pulling force

The formula for the y - component of a force \(F\) at an angle \(\theta\) is \(F_y = F\sin\theta\). Given \(F = 21.98\space N\) and \(\theta = 57.29^{\circ}\), \(\sin(57.29^{\circ})\approx0.84\), so \(F_y=21.98\times0.84 = 18.46\space N\) (again, the original hand - written answer used the wrong value for the force magnitude in the formula).

Step3: Calculate the weight of the sled

The formula for weight is \(F_w=mg\). Given \(m = 40.17\space kg\) and \(g = 9.8\space m/s^{2}\) (assuming standard gravity, if we take \(g = 10\space m/s^{2}\) as in the hand - written answer, \(F_w=40.17\times10=401.7\space N\)).

Step4: Calculate the normal force

In the y - direction, \(\sum F_y = N+F_y - mg=0\) (since there is no acceleration in the y - direction). So \(N=mg - F_y\). If \(mg = 401.7\space N\) (from step 3 with \(g = 10\space m/s^{2}\)) and \(F_y = 18.46\space N\) (from step 2), then \(N=401.7- 18.46=383.24\space N\) (the original hand - written answer used an incorrect formula \(F = ma\) for the normal force, the normal force is found from the equilibrium in the y - direction).

Step5: Calculate the acceleration of the sled

Using Newton's second law \(F = ma\), in the x - direction \(F = F_x\) (assuming no friction). Given \(F_x = 11.87\space N\) (from step 1) and \(m = 40.17\space kg\), \(a=\frac{F_x}{m}=\frac{11.87}{40.17}\approx0.296\space m/s^{2}\) (if we use the x - component calculation with \(F = 21.98\space N\) correctly, the original hand - written answer had \(a=\frac{21.98}{40.17}\approx0.55\space m/s^{2}\) but used the wrong component formula in step 1).

Answer:

(a) \(11.87\space N\)
(b) \(18.46\space N\)
(c) \(401.7\space N\) (if \(g = 10\space m/s^{2}\))
(d) \(383.24\space N\) (if \(g = 10\space m/s^{2}\))
(e) \(0.296\space m/s^{2}\) (using the correct x - component) or \(0.55\space m/s^{2}\) (if we assume the force magnitude was mis - written as \(40.17\) instead of \(21.98\) in the component calculation in the original problem setup)