A 21 kg block slides down a frictionless slope which is at angle θ = 25◦ . Starting from rest, the time to slide down is t = 1.51 s. The acceleration of gravity is 9.8 m/s 2 . 21 kg s 25◦ What total distance s did the block slide? Answer in units of

Answer :

Answer:

Explanation:

Given

mass of block=21 kg

inclination [tex]\theta =25^{\circ}[/tex]

t=1.51 s

total acceleration [tex]a=g\sin \theta =9.8\times \sin (25)=4.14 m/s^2[/tex]

[tex]v=u+at[/tex]

[tex]v=0+4.14\times 1.51=6.25 m/s[/tex]

[tex]v^2-u^2=2 as [/tex]

[tex]6.25^2-0=2\times 4.14\times s[/tex]

[tex]s=4.72 m[/tex]

Answer:

s  = 4.71 m

Explanation:

given,

mass of the block = 21 kg

angle  θ = 25◦                        

time to slide  = 1.51 s

acceleration = 9.8 m/s²

distance = ?                                    

acceleration = a sin θ                          

                     = 9.8 sin 25° = 4.14 m/s²

[tex]s = ut + \dfrac{1}{2}at^2[/tex]

[tex]s = \dfrac{1}{2}\times 4.14\times 1.51^2[/tex]

s  = 4.71 m

hence, the distance traveled by the block = 4.71 m

Other Questions