A 20 kg cart with frictionless bearings is initially at rest on a horizontal tabletop. The cart is connected by a light string to a hanging 5 kg block. The string passes over an ideal pulley. The system is released from rest and both objects have an acceleration of magnitude a 5 kg
(a) Draw the forces (NOT components) that act on each object.
(b) Calculate the acceleration ay of the system of masses once the system is released.

Respuesta :

Answer:

[tex]a=1.96m/s^2[/tex]

Explanation:

From the question we are told that

Mass of cart =20kg

Mass of block =5kg

Magnitude of acceleration M_a= 5 kg

Generally Force on the 20kg mass in the [tex]\bar X[/tex] is mathematically given as

[tex]\sum F=ma\\t=(20kg)a[/tex]

Generally Force on the 2kg in the[tex]\bar Y[/tex] mass is mathematically given as

[tex]\sum F=ma\\(5kg)(9.8m/s^2)-(5kg)a[/tex]

Therefore

[tex](20kg)a=(5kg)(9.8m/s^2)-(5kg)a[/tex]

[tex]a=\frac{5kg)(9.8m/s^2)-(5kg)}{25}[/tex]

[tex]a=1.96m/s^2[/tex]

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