What is the acceleration along the ground of a 10 kg wagon when it is pulled with a force of 44 n at an angle of 35° above the horizontal? 0.23 m/s2 0.28 m/s2 2.52 m/s2 3.60 m/s2

Via a free body diagram we obtain that the x force component is and the y force component is

In this case and , since we want to know the acceleration along the ground we only need to use the x-force component force and so by substituting into our x-force component we obtain:

a=3.60 m/s^2

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer