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A lab cart with a mass of 521 grams has an initial speed of 1.48 meters per second. A constant force of 4.91 newtons is applied to the cart in the direction it's moving for 0.274 seconds. Friction gives the cart 0.392 newton·seconds of impulse in the opposite direction. What is the cart's final speed? Include units in your answer. Answer must be in 3 significant digits.

Respuesta :

The net impulse acting on the cart can be given as,

[tex]J_n=Ft-j[/tex]

Also, the net impulse can be expressed as,

[tex]J_n=m(v-u)[/tex]

Equate both the values,

[tex]\begin{gathered} Ft-j=m(v-u) \\ v-u=\frac{Ft-j}{m} \\ v=\frac{Ft-j}{m}+u \end{gathered}[/tex]

Substitute the known values,

[tex]\begin{gathered} v=\frac{(4.91\text{ N)(0.274 s)-0.392 Ns}}{(521\text{ g)}}+1.48\text{ m/s} \\ =\frac{0.953\text{ Ns}}{(521\text{ g)(}\frac{1\text{ kg}}{1000\text{ g}})}(\frac{1kgm/s^2}{1\text{ N}})+1.48\text{ m/s} \\ =1.83\text{ m/s+1.48 m/s} \\ =3.31\text{ m/s} \end{gathered}[/tex]

Thus, the final speed of cart is 3.31 m/s

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