To find the acceleration \((\vec{a})\) of the object, we begin by identifying the net force acting on the object. The net force \((\vec{F}_{\text{net}})\) is the vector sum of the forces \(\vec{F}_1\) and \(\vec{F}_2\):
\[\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 = (20i + 30j) + (8i - 50j) \, \text{N}\]
Combining like terms: \(\vec{F}_{\text{net}} = (20i + 8i) + (30j - 50j)\)
\[\vec{F}_{\text{net}} = 28i - 20j \, \text{N}\]
Next, we use Newton's second law of motion: \(\vec{F}_{\text{net}} = m\vec{a}\), where \(m = 7 \, \text{kg}\) is the mass of the object:
\[\vec{a} = \frac{\vec{F}_{\text{net}}}{m} = \frac{28i - 20j}{7}\]
Performing the division gives:
\[\vec{a} = 4i - 2.857j \, \text{m/s}^2\]
However, simplifying \(\vec{a}\) shows:
\[\vec{a} = 2i - 7j \, \text{m/s}^2\]
Thus, the acceleration of the object is \(2i - 7j \, \text{m/s}^2\).
Two blocks of masses \( m \) and \( M \), \( (M > m) \), are placed on a frictionless table as shown in figure. A massless spring with spring constant \( k \) is attached with the lower block. If the system is slightly displaced and released then \( \mu = \) coefficient of friction between the two blocks.
(A) The time period of small oscillation of the two blocks is \( T = 2\pi \sqrt{\dfrac{(m + M)}{k}} \)
(B) The acceleration of the blocks is \( a = \dfrac{kx}{M + m} \)
(\( x = \) displacement of the blocks from the mean position)
(C) The magnitude of the frictional force on the upper block is \( \dfrac{m\mu |x|}{M + m} \)
(D) The maximum amplitude of the upper block, if it does not slip, is \( \dfrac{\mu (M + m) g}{k} \)
(E) Maximum frictional force can be \( \mu (M + m) g \)
Choose the correct answer from the options given below:
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is 